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• Use an Ubuntu Live CD to Securely Wipe Your PC’s Hard Drive

  - by Trevor Bekolay

  Deleting files or quickly formatting a drive isn’t enough for sensitive personal information. We’ll show you how to get rid of it for good using a Ubuntu Live CD. When you delete a file in Windows, Ubuntu, or any other operating system, it doesn’t actually destroy the data stored on your hard drive, it just marks that data as “deleted.” If you overwrite it later, then that data is generally unrecoverable, but if the operating system don’t happen to overwrite it, then your data is still stored on your hard drive, recoverable by anyone who has the right software. By securely delete files or entire hard drives, your data will be gone for good. Note: Modern hard drives are extremely sophisticated, as are the experts who recover data for a living. There is no guarantee that the methods covered in this article will make your data completely unrecoverable; however, they will make your data unrecoverable to the majority of recovery methods, and all methods that are readily available to the general public. Shred individual files Most of the data stored on your hard drive is harmless, and doesn’t reveal anything about you. If there are just a few files that you know you don’t want someone else to see, then the easiest way to get rid of them is a built-in Linux utility called shred. Open a terminal window by clicking on Applications at the top-left of the screen, then expanding the Accessories menu and clicking on Terminal. Navigate to the file that you want to delete using cd to change directories and ls to list the files and folders in the current directory. As an example, we’ve got a file called BankInfo.txt on a Windows NTFS-formatted hard drive. We want to delete it securely, so we’ll call shred by entering the following in the terminal window: shred <file> which is, in our example: shred BankInfo.txt Notice that our BankInfo.txt file still exists, even though we’ve shredded it. A quick look at the contents of BankInfo.txt make it obvious that the file has indeed been securely overwritten. We can use some command-line arguments to make shred delete the file from the hard drive as well. We can also be extra-careful about the shredding process by upping the number of times shred overwrites the original file. To do this, in the terminal, type in: shred –remove –iterations=<num> <file> By default, shred overwrites the file 25 times. We’ll double this, giving us the following command: shred –remove –iterations=50 BankInfo.txt BankInfo.txt has now been securely wiped on the physical disk, and also no longer shows up in the directory listing. Repeat this process for any sensitive files on your hard drive! Wipe entire hard drives If you’re disposing of an old hard drive, or giving it to someone else, then you might instead want to wipe your entire hard drive. shred can be invoked on hard drives, but on modern file systems, the shred process may be reversible. We’ll use the program wipe to securely delete all of the data on a hard drive. Unlike shred, wipe is not included in Ubuntu by default, so we have to install it. Open up the Synaptic Package Manager by clicking on System in the top-left corner of the screen, then expanding the Administration folder and clicking on Synaptic Package Manager. wipe is part of the Universe repository, which is not enabled by default. We’ll enable it by clicking on Settings > Repositories in the Synaptic Package Manager window. Check the checkbox next to “Community-maintained Open Source software (universe)”. Click Close. You’ll need to reload Synaptic’s package list. Click on the Reload button in the main Synaptic Package Manager window. Once the package list has been reloaded, the text over the search field will change to “Rebuilding search index”. Wait until it reads “Quick search,” and then type “wipe” into the search field. The wipe package should come up, along with some other packages that perform similar functions. Click on the checkbox to the left of the label “wipe” and select “Mark for Installation”. Click on the Apply button to start the installation process. Click the Apply button on the Summary window that pops up. Once the installation is done, click the Close button and close the Synaptic Package Manager window. Open a terminal window by clicking on Applications in the top-left of the screen, then Accessories > Terminal. You need to figure our the correct hard drive to wipe. If you wipe the wrong hard drive, that data will not be recoverable, so exercise caution! In the terminal window, type in: sudo fdisk -l A list of your hard drives will show up. A few factors will help you identify the right hard drive. One is the file system, found in the System column of  the list – Windows hard drives are usually formatted as NTFS (which shows up as HPFS/NTFS). Another good identifier is the size of the hard drive, which appears after its identifier (highlighted in the following screenshot). In our case, the hard drive we want to wipe is only around 1 GB large, and is formatted as NTFS. We make a note of the label found under the the Device column heading. If you have multiple partitions on this hard drive, then there will be more than one device in this list. The wipe developers recommend wiping each partition separately. To start the wiping process, type the following into the terminal: sudo wipe <device label> In our case, this is: sudo wipe /dev/sda1 Again, exercise caution – this is the point of no return! Your hard drive will be completely wiped. It may take some time to complete, depending on the size of the drive you’re wiping. Conclusion If you have sensitive information on your hard drive – and chances are you probably do – then it’s a good idea to securely delete sensitive files before you give away or dispose of your hard drive. The most secure way to delete your data is with a few swings of a hammer, but shred and wipe from a Ubuntu Live CD is a good alternative! Similar Articles Productive Geek Tips Reset Your Ubuntu Password Easily from the Live CDScan a Windows PC for Viruses from a Ubuntu Live CDRecover Deleted Files on an NTFS Hard Drive from a Ubuntu Live CDCreate a Bootable Ubuntu 9.10 USB Flash DriveCreate a Bootable Ubuntu USB Flash Drive the Easy Way TouchFreeze Alternative in AutoHotkey The Icy Undertow Desktop Windows Home Server – Backup to LAN The Clear & Clean Desktop Use This Bookmarklet to Easily Get Albums Use AutoHotkey to Assign a Hotkey to a Specific Window Latest Software Reviews Tinyhacker Random Tips DVDFab 6 Revo Uninstaller Pro Registry Mechanic 9 for Windows PC Tools Internet Security Suite 2010 Office 2010 Product Guides Google Maps Place marks – Pizza, Guns or Strip Clubs Monitor Applications With Kiwi LocPDF is a Visual PDF Search Tool Download Free iPad Wallpapers at iPad Decor Get Your Delicious Bookmarks In Firefox’s Awesome Bar

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• What Every Developer Should Know About MSI Components

  - by Alois Kraus

  Hopefully nothing. But if you have to do more than simple XCopy deployment and you need to support updates, upgrades and perhaps side by side scenarios there is no way around MSI. You can create Msi files with a Visual Studio Setup project which is severely limited or you can use the Windows Installer Toolset. I cannot talk about WIX with my German colleagues because WIX has a very special meaning. It is funny to always use the long name when I talk about deployment possibilities. Alternatively you can buy commercial tools which help you to author Msi files but I am not sure how good they are. Given enough pain with existing solutions you can also learn the MSI Apis and create your own packaging solution. If I were you I would use either a commercial visual tool when you do easy deployments or use the free Windows Installer Toolset. Once you know the WIX schema you can create well formed wix xml files easily with any editor. Then you can “compile” from the wxs files your Msi package. Recently I had the “pleasure” to get my hands dirty with C++ (again) and the MSI technology. Installation is a complex topic but after several month of digging into arcane MSI issues I can safely say that there should exist an easier way to install and update files as today. I am not alone with this statement as John Robbins (creator of the cool tool Paraffin) states: “.. It's a brittle and scary API in Windows …”. To help other people struggling with installation issues I present you the advice I (and others) found useful and what will happen if you ignore this advice. What is a MSI file? A MSI file is basically a database with tables which reference each other to control how your un/installation should work. The basic idea is that you declare via these tables what you want to install and MSI controls the how to get your stuff onto or off your machine. Your “stuff” consists usually of files, registry keys, shortcuts and environment variables. Therefore the most important tables are File, Registry, Environment and Shortcut table which define what will be un/installed. The key to master MSI is that every resource (file, registry key ,…) is associated with a MSI component. The actual payload consists of compressed files in the CAB format which can either be embedded into the MSI file or reside beside the MSI file or in a subdirectory below it. To examine MSI files you need Orca a free MSI editor provided by MS. There is also another free editor called Super Orca which does support diffs between MSI and it does not lock the MSI files. But since Orca comes with a shell extension I tend to use only Orca because it is so easy to right click on a MSI file and open it with this tool. How Do I Install It? Double click it. This does work for fresh installations as well as major upgrades. Updates need to be installed via the command line via msiexec /i <msi> REINSTALL=ALL REINSTALLMODE=vomus   This tells the installer to reinstall all already installed features (new features will NOT be installed). The reinstallmode letters do force an overwrite of the old cached package in the %WINDIR%\Installer folder. All files, shortcuts and registry keys are redeployed if they are missing or need to be replaced with a newer version. When things did go really wrong and you want to overwrite everything unconditionally use REINSTALLMODE=vamus. How To Enable MSI Logs? You can download a MSI from Microsoft which installs some registry keys to enable full MSI logging. The log files can be found in your %TEMP% folder and are called MSIxxxx.log. Alternatively you can add to your msiexec command line the option msiexec …. /l*vx <LogFileName> Personally I find it rather strange that * does not mean full logging. To really get all logs I need to add v and x which is documented in the msiexec help but I still find this behavior unintuitive. What are MSI components? The whole MSI logic is bound to the concept of MSI components. Nearly every msi table has a Component column which binds an installable resource to a component. Below are the screenshots of the FeatureComponents and Component table of an example MSI. The Feature table defines basically the feature hierarchy.  To find out what belongs to a feature you need to look at the FeatureComponents table where for each feature the components are listed which will be installed when a feature is installed. The MSI components are defined in the  Component table. This table has as first column the component name and as second column the component id which is a GUID. All resources you want to install belong to a MSI component. Therefore nearly all MSI tables have a Component_ column which contains the component name. If you look e.g. a the File table you see that every file belongs to a component which is true for all other tables which install resources. The component table is the glue between all other tables which contain the resources you want to install. So far so easy. Why is MSI then so complex? Most MSI problems arise from the fact that you did violate a MSI component rule in one or the other way. When you install a feature the reference count for all components belonging to this feature will increase by one. If your component is installed by more than one feature it will get a higher refcount. When you uninstall a feature its refcount will drop by one. Interesting things happen if the component reference count reaches zero: Then all associated resources will be deleted. That looks like a reasonable thing and it is. What it makes complex are the strange component rules you have to follow. Below are some important component rules from the Tao of the Windows Installer … Rule 16: Follow Component Rules Components are a very important part of the Installer technology. They are the means whereby the Installer manages the resources that make up your application. The SDK provides the following guidelines for creating components in your package: Never create two components that install a resource under the same name and target location. If a resource must be duplicated in multiple components, change its name or target location in each component. This rule should be applied across applications, products, product versions, and companies. Two components must not have the same key path file. This is a consequence of the previous rule. The key path value points to a particular file or folder belonging to the component that the installer uses to detect the component. If two components had the same key path file, the installer would be unable to distinguish which component is installed. Two components however may share a key path folder. Do not create a version of a component that is incompatible with all previous versions of the component. This rule should be applied across applications, products, product versions, and companies. Do not create components containing resources that will need to be installed into more than one directory on the user’s system. The installer installs all of the resources in a component into the same directory. It is not possible to install some resources into subdirectories. Do not include more than one COM server per component. If a component contains a COM server, this must be the key path for the component. Do not specify more than one file per component as a target for the Start menu or a Desktop shortcut. … And these rules do not even talk about component ids, update packages and upgrades which you need to understand as well. Lets suppose you install two MSIs (MSI1 and MSI2) which have the same ComponentId but different component names. Both do install the same file. What will happen when you uninstall MSI2?   Hm the file should stay there. But the component names are different. Yes and yes. But MSI uses not use the component name as key for the refcount. Instead the ComponentId column of the Component table which contains a GUID is used as identifier under which the refcount is stored. The components Comp1 and Comp2 are identical from the MSI perspective. After the installation of both MSIs the Component with the Id {100000….} has a refcount of two. After uninstallation of one MSI there is still a refcount of one which drops to zero just as expected when we uninstall the last msi. Then the file which was the same for both MSIs is deleted. You should remember that MSI keeps a refcount across MSIs for components with the same component id. MSI does manage components not the resources you did install. The resources associated with a component are then and only then deleted when the refcount of the component reaches zero.   The dependencies between features, components and resources can be described as relations. m,k are numbers >= 1, n can be 0. Inside a MSI the following relations are valid Feature    1  –> n Components Component    1 –> m Features Component      1  –>  k Resources These relations express that one feature can install several components and features can share components between them. Every (meaningful) component will install at least one resource which means that its name (primary key to stay in database speak) does occur in some other table in the Component column as value which installs some resource. Lets make it clear with an example. We want to install with the feature MainFeature some files a registry key and a shortcut. We can then create components Comp1..3 which are referenced by the resources defined in the corresponding tables.   Feature Component Registry File Shortcuts MainFeature Comp1 RegistryKey1     MainFeature Comp2   File.txt   MainFeature Comp3   File2.txt Shortcut to File2.txt   It is illegal that the same resource is part of more than one component since this would break the refcount mechanism. Lets illustrate this:            Feature ComponentId Resource Reference Count Feature1 {1000-…} File1.txt 1 Feature2 {2000-….} File1.txt 1 The installation part works well but what happens when you uninstall Feature2? Component {20000…} gets a refcount of zero where MSI deletes all resources belonging to this component. In this case File1.txt will be deleted. But Feature1 still has another component {10000…} with a refcount of one which means that the file was deleted too early. You just have ruined your installation. To fix it you then need to click on the Repair button under Add/Remove Programs to let MSI reinstall any missing registry keys, files or shortcuts. The vigilant reader might has noticed that there is more in the Component table. Beside its name and GUID it has also an installation directory, attributes and a KeyPath. The KeyPath is a reference to a file or registry key which is used to detect if the component is already installed. This becomes important when you repair or uninstall a component. To find out if the component is already installed MSI checks if the registry key or file referenced by the KeyPath property does exist. When it does not exist it assumes that it was either already uninstalled (can lead to problems during uninstall) or that it is already installed and all is fine. Why is this detail so important? Lets put all files into one component. The KeyPath should be then one of the files of your component to check if it was installed or not. When your installation becomes corrupt because a file was deleted you cannot repair it with the Repair button under Add/Remove Programs because MSI checks the component integrity via the Resource referenced by its KeyPath. As long as you did not delete the KeyPath file MSI thinks all resources with your component are installed and never executes any repair action. You get even more trouble when you try to remove files during an upgrade (you cannot remove files during an update) from your super component which contains all files. The only way out and therefore best practice is to assign for every resource you want to install an extra component. This ensures painless updatability and repairs and you have much less effort to remove specific files during an upgrade. In effect you get this best practice relation Feature 1  –> n Components Component   1  –>  1 Resources MSI Component Rules Rule 1 – One component per resource Every resource you want to install (file, registry key, value, environment value, shortcut, directory, …) must get its own component which does never change between versions as long as the install location is the same. Penalty If you add more than one resources to a component you will break the repair capability of MSI because the KeyPath is used to check if the component needs repair. MSI ComponentId Files MSI 1.0 {1000} File1-5 MSI 2.0 {2000} File2-5 You want to remove File1 in version 2.0 of your MSI. Since you want to keep the other files you create a new component and add them there. MSI will delete all files if the component refcount of {1000} drops to zero. The files you want to keep are added to the new component {2000}. Ok that does work if your upgrade does uninstall the old MSI first. This will cause the refcount of all previously installed components to reach zero which means that all files present in version 1.0 are deleted. But there is a faster way to perform your upgrade by first installing your new MSI and then remove the old one.  If you choose this upgrade path then you will loose File1-5 after your upgrade and not only File1 as intended by your new component design.   Rule 2 – Only add, never remove resources from a component If you did follow rule 1 you will not need Rule 2. You can add in a patch more resources to one component. That is ok. But you can never remove anything from it. There are tricky ways around that but I do not want to encourage bad component design. Penalty Lets assume you have 2 MSI files which install under the same component one file   MSI1 MSI2 {1000} - ComponentId {1000} – ComponentId File1.txt File2.txt   When you install and uninstall both MSIs you will end up with an installation where either File1 or File2 will be left. Why? It seems that MSI does not store the resources associated with each component in its internal database. Instead Windows will simply query the MSI that is currently uninstalled for all resources belonging to this component. Since it will find only one file and not two it will only uninstall one file. That is the main reason why you never can remove resources from a component!   Rule 3 Never Remove A Component From an Update MSI. This is the same as if you change the GUID of a component by accident for your new update package. The resulting update package will not contain all components from the previously installed package. Penalty When you remove a component from a feature MSI will set the feature state during update to Advertised and log a warning message into its log file when you did enable MSI logging. SELMGR: ComponentId '{2DCEA1BA-3E27-E222-484C-D0D66AEA4F62}' is registered to feature 'xxxxxxx, but is not present in the Component table.  Removal of components from a feature is not supported! MSI (c) (24:44) [07:53:13:436]: SELMGR: Removal of a component from a feature is not supported Advertised means that MSI treats all components of this feature as not installed. As a consequence during uninstall nothing will be removed since it is not installed! This is not only bad because uninstall does no longer work but this feature will also not get the required patches. All other features which have followed component versioning rules for update packages will be updated but the one faulty feature will not. This results in very hard to find bugs why an update was only partially successful. Things got better with Windows Installer 4.5 but you cannot rely on that nobody will use an older installer. It is a good idea to add to your update msiexec call MSIENFORCEUPGRADECOMPONENTRULES=1 which will abort the installation if you did violate this rule.

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• My files disappeared from the UbuntuOne synced folder

  - by Junji

  I set up an UbuntuOne account on PC1 (Ubuntu 10.10) and the same account on PC2 (Ubuntu 10.04). I did the following: Created file named maverick.txt in PC1's ~/Ubuntu One/log Created file named venus.txt in PC2's ~/Ubuntu One/log Bot files appeared in one.ubuntu.com A few hours later, those two files are disappeared from PC1's Ubuntu One/log PC2's Ubuntu One/log one.ubuntu.com So, my files are gone forever. Why did this happen? Is there any way to recover those files?

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• Zenoss Setup for Windows Servers

  - by Jay Fox

  Recently I was saddled with standing up Zenoss for our enterprise.  We're running about 1200 servers, so manually touching each box was not an option.  We use LANDesk for a lot of automated installs and patching - more about that later.The steps below may not necessarily have to be completed in this order - it's just the way I did it.STEP ONE:Setup a standard AD user.  We want to do this so there's minimal security exposure.  Call the account what ever you want "domain/zenoss" for our examples.***********************************************************STEP TWO:Make the following local groups accessible by your zenoss account.Distributed COM UsersPerformance Monitor UsersEvent Log Readers (which doesn't exist on pre-2008 machines)Here's the Powershell script I used to setup access to these local groups:# Created to add Active Directory account to local groups# Must be run from elevated prompt, with permissions on the remote machine(s).# Create txt file should contain the names of the machines that need the account added, one per line.# Script will process machines line by line.foreach($i in (gc c:\tmp\computers.txt)){# Add the user to the first group$objUser=[ADSI]("WinNT://domain/zenoss")$objGroup=[ADSI]("WinNT://$i/Distributed COM Users")$objGroup.PSBase.Invoke("Add",$objUser.PSBase.Path)# Add the user to the second group$objUser=[ADSI]("WinNT://domain/zenoss")$objGroup=[ADSI]("WinNT://$i/Performance Monitor Users")$objGroup.PSBase.Invoke("Add",$objUser.PSBase.Path)# Add the user to the third group - Group doesn't exist on < Server 2008#$objUser=[ADSI]("WinNT://domain/zenoss")#$objGroup=[ADSI]("WinNT://$i/Event Log Readers")#$objGroup.PSBase.Invoke("Add",$objUser.PSBase.Path)}**********************************************************STEP THREE:Setup security on the machines namespace so our domain/zenoss account can access itThe default namespace for zenoss is:  root/cimv2Here's the Powershell script:#Grant account defined below (line 11) access to WMI Namespace#Has to be run as account with permissions on remote machinefunction get-sid{Param ($DSIdentity)$ID = new-object System.Security.Principal.NTAccount($DSIdentity)return $ID.Translate( [System.Security.Principal.SecurityIdentifier] ).toString()}$sid = get-sid "domain\zenoss"$SDDL = "A;;CCWP;;;$sid" $DCOMSDDL = "A;;CCDCRP;;;$sid"$computers = Get-Content "c:\tmp\computers.txt"foreach ($strcomputer in $computers){    $Reg = [WMIClass]"\\$strcomputer\root\default:StdRegProv"    $DCOM = $Reg.GetBinaryValue(2147483650,"software\microsoft\ole","MachineLaunchRestriction").uValue    $security = Get-WmiObject -ComputerName $strcomputer -Namespace root/cimv2 -Class __SystemSecurity    $converter = new-object system.management.ManagementClass Win32_SecurityDescriptorHelper    $binarySD = @($null)    $result = $security.PsBase.InvokeMethod("GetSD",$binarySD)    $outsddl = $converter.BinarySDToSDDL($binarySD[0])    $outDCOMSDDL = $converter.BinarySDToSDDL($DCOM)    $newSDDL = $outsddl.SDDL += "(" + $SDDL + ")"    $newDCOMSDDL = $outDCOMSDDL.SDDL += "(" + $DCOMSDDL + ")"    $WMIbinarySD = $converter.SDDLToBinarySD($newSDDL)    $WMIconvertedPermissions = ,$WMIbinarySD.BinarySD    $DCOMbinarySD = $converter.SDDLToBinarySD($newDCOMSDDL)    $DCOMconvertedPermissions = ,$DCOMbinarySD.BinarySD    $result = $security.PsBase.InvokeMethod("SetSD",$WMIconvertedPermissions)     $result = $Reg.SetBinaryValue(2147483650,"software\microsoft\ole","MachineLaunchRestriction", $DCOMbinarySD.binarySD)}***********************************************************STEP FOUR:Get the SID for our zenoss account.Powershell#Provide AD User get SID$objUser = New-Object System.Security.Principal.NTAccount("domain", "zenoss") $strSID = $objUser.Translate([System.Security.Principal.SecurityIdentifier]) $strSID.Value******************************************************************STEP FIVE:Modify the Service Control Manager to allow access to the zenoss AD account.This command can be run from an elevated command line, or through Powershellsc sdset scmanager "D:(A;;CC;;;AU)(A;;CCLCRPRC;;;IU)(A;;CCLCRPRC;;;SU)(A;;CCLCRPWPRC;;;SY)(A;;KA;;;BA)(A;;CCLCRPRC;;;PUT_YOUR_SID_HERE_FROM STEP_FOUR)S:(AU;FA;KA;;;WD)(AU;OIIOFA;GA;;;WD)"******************************************************************In step two the script plows through a txt file that processes each computer listed on each line.  For the other scripts I ran them on each machine using LANDesk.  You can probably edit those scripts to process a text file as well.That's what got me off the ground monitoring the machines using Zenoss.  Hopefully this is helpful for you.  Watch the line breaks when copy the scripts.

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• How can I create an SPF record on my 1and1.com hosted domain?

  - by tnorthcutt

  Emails from my domain (hosted at 1and1, and using Google Apps Premier edition) have sporadically been going to recipients' spam folders lately. I did some research, tested, and found out that I do not have an SPF record for my domain. According to this Google Support page, I need to create one. Following the steps on that page is easy, until I get to #3: Create a TXT record containing this text: v=spf1 include:_spf.google.com ~all I see no way to create a "TXT record". Here is a screenshot of the admin panel:

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• How can I set up the Trusty Desktop ISO for preseeding?

  - by user288087

  With Precise, I used to create a USB stick using the so-called "Alternate" installer DVD with my own preseed.txt, update isolinux/txt.cfg with a new menu item that would pass my preseed file on the command line, and watch as the installer happily did all the right things to create a Ubuntu install on the hard drive of the machine I booted. With Trusty, the 'alternate' image has disappeared. Is it still possible to preseed a Trusty install, and if so, how do I get the d-i/ubiquity system to use my preseed file?

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• Squid on an Azure VM

  - by LantisGaius

  I can't get it to work. Here's exactly what I did: Create a new Azure VM, Windows Server 2012. RDP to the new VM Download & Extract Squid for Windows (2.7.STABLE8) Rename the conf files (squid, mime & cachemgr) Add the following lines on the end of squid.conf auth_param basic program c:/squid/libexec/ncsa_auth.exe c:/squid/etc/passwd.txt auth_param basic children 5 auth_param basic realm Welcome to http://abcde.fg Squid Proxy! auth_param basic credentialsttl 12 hours auth_param basic casesensitive off acl ncsa_users proxy_auth REQUIRED http_access allow ncsa_users Use http://www.htaccesstools.com/htpasswd-generator-windows/ to create passwd.txt Test passwd.txt via c:/squid/libexec/ncsa_auth.exe c:/squid/etc/passwd.txt (success) squid -z squid -i net start squid (No errors so far). go to https://manage.windowsazure.com, Virtual Machines - myVM - Endpoints Add Endpoint: Name: Squid Protocol: TCP Public Port: 80 Private Port: 3128 That's it. Unfortunately, it doesn't work. I think I screwed something up at the endpoint? I'm not sure.. help? EDIT: I'm testing it via Firefox - Options - Advanced - Network, and the exact error is "The Proxy Server is refusing connections." I'm using my DNS as the Proxy server "abcdef.cloudapp.net" and port 80 (since that's my public endpoint).

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• sequential SSH command execution not working in Ubuntu/Bash

  - by kumar

  My requirement is I will have a set of commands that needs to be executed in a text file. My Shell script has to read each command, execute and store the results in a separate file. Here is the snippet which does the above requirement. while read command do echo 'Command :' $command >> "$OUTPUT_FILE" redirect_pos=`expr index "$command" '>>'` if [ `expr index "$command" '>>'` != 0 ];then redirect_fn "$redirect_pos" "$command"; else $command state=$? if [ $state != 0 ];then echo "command failed." >> "$OUTPUT_FILE" else echo "executed successfully." >> "$OUTPUT_FILE" fi fi echo >> "$OUTPUT_FILE" done < "$INPUT_FILE" Sample Commands.txt will be like this ... tar -rvf /var/tmp/logs.tar -C /var/tmp/ Commands_log.txt gzip /var/tmp/logs.tar rm -f /var/tmp/list.txt This is working fine for commands which needs to be executed in local machine. But When I am trying to execute the following ssh commands only the 1st command getting executed. Here are the some of the ssh commands added in my text file. ssh uname@hostname1 tar -rvf /var/tmp/logs.tar -C /var/tmp/ Commands_log.txt ssh uname@hostname2 gzip /var/tmp/logs.tar ssh .. etc When I am executing this in cli it is working fine. Could anybody help me in this?

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• Do you need to advertise an AFP service via Avahi for an Ubuntu Server to show up in OSX Finder?

  - by James

  I am only advertising an NFS share plus the "model", and I don't want to install extra services on the Server unless I have to, ie netatalk, as it is used solely for NFS exports. Currently there is no entry in Finder under "Shared" with below config of Avahi. serveradmin@FILESERVER:/etc/avahi/services$ cat nfs.service <?xml version="1.0" standalone='no'?><!--*-nxml-*--> <!DOCTYPE service-group SYSTEM "avahi-service.dtd"> <service-group> <name replace-wildcards="yes">%h</name> <service> <type>_nfs._tcp</type> <port>2049</port> <txt-record>path=/Volumes/StoragePool</txt-record> </service> <service> <type>_device-info._tcp</type> <port>0</port> <txt-record>model=Xserve</txt-record> </service> </service-group> Server: Ubuntu 12.04.01 x64 Clients: OSX 10.6.8 , 10.7.5, 10.8.2 The goal is to advertise that NFS share, then assign a really old Model code of Mac like a Powermac and switch out the icon for a more "LinuxServer-y" one. Plus allow users to connect to NFS in a manner they are familiar with like our other Xserve servers. I think Avahi is working in general as if I do: nfs://FILESERVER.local/Volumes/StoragePool it will connect fine. Any ideas?

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• PsExec and Remote Environment Variables, Logging, Etc.

  - by alharaka

  When I run PsExec on a remote computer, I always fall short of what I want. What I would like ideally in most situations is a) a log on an admin server where each individual log has the name of each the remote computer it was generated from (e.g. COMPNAME1.log, COMPNAME2.log, etc.) or b) a log file on each remote computer with whatever name I specify. When I try scenario (a), I use the following command. %SystemDrive%\path\to\psexec.exe @listofcomputers.txt -u DOMAIN\username cmd /c echo TEST >> \\server.company.tld\share\%computername%.log Problem is that it never works. All the computers just write to the log where %computername% is just the computer I execute PsExec from in my office. What I want are unique logs for each computer specific in the listofcomputers.txt that will correctly use the hostname from the remote environment variable without issue. Is that even possible? It does not seem to work for me. I tried this, and the syntax is clearly wrong. %SystemDrive%\path\to\psexec.exe @listofcomputers.txt -u DOMAIN\username "cmd /c echo TEST >> \\server.company.tld\share\%computername%.log" PsExec just fails saying the system file cannot be found (read: syntax fail). As for scenario (b), it appears to be a variation of a similar problem. When I run a command like this, it does not work. %SystemDrive%\path\to\psexec.exe @listofcomputers.txt -u DOMAIN\username "cmd /c echo %computername% >> \\server.company.tld\share\aggregated.log" Is there something I do not understand about remote path and environment variables with PsExec on the cmd.exe console (I have not even tried the dreaded PowerShell yet). I know such things work in a batch file (cmd /c \\server.company.tld\share\runthis.bat), but is there a reason it will not work when executing commands as arguments? I always need this, and can never get it!

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• DNS zone file SPF configuration to support sending mail from multiple servers and gmail

  - by Tauren

  I want to configure SPF on a domain to allow mail to be sent from: the x.com website server (x.com and www.x.com - both at same IP) it's MX servers (smtp.x.com, mx.x.com, mail.x.com) another server that isn't listed as an MX server (somehost.x.com) via gmail using an account that has authenticated use of [email protected] Will this zone file work? If not, what are the problems with it? $ttl 38400 @ IN SOA ns1.x.com. hostmaster.x.com. ( 201003092 ; serial 8H ; refresh 15M ; retry 1W ; expire 1H ) ; minimum @ NS ns1.x.com. @ NS ns2.x.com. @ MX 10 mx.x.com. @ MX 20 smtp.x.com. @ MX 30 mailhost.x.com. ; SPF records @ IN TXT "v=spf1 a mx a:somehost.x.com include:_spf.google.com ~all" mx IN TXT "v=spf1 a -all" smtp IN TXT "v=spf1 a -all" mailhost IN TXT "v=spf1 a -all" Questions: Is _spf.google.com the right thing to include for gmail.com, or is it only for Google Hosted Apps? If only for Google Apps, what should I include to send from gmail.com? If mail shouldn't be sent from anywhere else, is it safe to use -all instead of ~all? Does it make sense to add specific SPF records for each of the mail servers? Any other problems with the zone file? I want to confirm these things before making changes to my zone file. The file has SPF configured basically the same now, just without google.com and somehost, but I want to make sure I won't break things when I change it.

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• Adding License to VMware Server 2 via scripting command?

  - by andyt25

  Hi all, I recently discovered the vimsvc/license command in vmware-vim-cmd and was trying to use that to automatically add my license key to a fresh vmware installation. vmware-vim-cmd -H hostip -O portnumber vimsvc/license --source file '/path/to/plaintext-file-that-contains-my-license-key.txt' plaintext-file-that-contains-my-license-key.txt contains my key in XXXXX-XXXXX-XXXXX-XXXXX format, I've also tried it with an extra carriage return at the end. Adding the key that way doesn't work, however. I always get the following error message: [200] Reading local file: /path/to/plaintext-file-that-contains-my-license-key.txt [200] Size of file is 24 bytes. returned were XXXXX-XXXXX-XXXXX-XXXXX [200] Changing license source to: file:/path/to/plaintext-file-that-contains-my-license-key.txt [500] Caught unexpected exception Type: N5Vmomi5Fault17NotEnoughLicenses9ExceptionE what() =vmodl.fault.NotEnoughLicenses GetMsg() = There are not enough licenses installed to perform the operation. It's kinda silly to require a license to be able to add a license, don't you think? ;-) So how do I go about and add the key via script? I would like to avoid any interaction as I have the rest of the install fully scripted and non-interactive. Kind Regards, Stefan

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• Usage of putty in command line from Hudson

  - by kij

  Hi, I'm trying to use putty in command line from an hudson job. The command is the following one: putty -ssh -2 -P 22 USERNAME@SERVER_ADDR -pw PASS -m command.txt Where 'command.txt' is a shell script to execute in the server through SSH. If i launch this command from the Window command prompt, it works, the shell script is executed on the server machine. If i launch a build of the hudson job configured with this batch command, it doesn't work. The build is running... and running... and running.. without doing anything, and i have to stop it manually. So my question is: Is it possible to launch an external programm (i.e. putty) from an hudson job ? ps: i tried SSH plugin but... not a really good plugin (pre/post build, fail status of the commands launched not caught by hudson, etc.) Thanks in advance for your help. Best regards. kij EDIT: These are the build logs: [workspace] $ cmd /c call C:\WINDOWS\TEMP\hudson7429256014041663539.bat C:\Hudson\jobs\Artifact deployer\workspace>putty -ssh -2 -P 22 USER@SERV_ADD -pw PASS -m com.txt Le build a été annulé Finished: ABORTED And the Hudson.err.log file at the same time (after a stop): 3 juin 2010 18:27:28 hudson.model.Run run INFO: Artifact deployer #6 aborted java.lang.InterruptedException at java.lang.ProcessImpl.waitFor(Native Method) at hudson.Proc$LocalProc.join(Proc.java:179) at hudson.Launcher$ProcStarter.join(Launcher.java:278) at hudson.tasks.CommandInterpreter.perform(CommandInterpreter.java:83) at hudson.tasks.CommandInterpreter.perform(CommandInterpreter.java:58) at hudson.tasks.BuildStepMonitor$1.perform(BuildStepMonitor.java:19) at hudson.model.AbstractBuild$AbstractRunner.perform(AbstractBuild.java:601) at hudson.model.Build$RunnerImpl.build(Build.java:174) at hudson.model.Build$RunnerImpl.doRun(Build.java:138) at hudson.model.AbstractBuild$AbstractRunner.run(AbstractBuild.java:416) at hudson.model.Run.run(Run.java:1241) at hudson.model.FreeStyleBuild.run(FreeStyleBuild.java:46) at hudson.model.ResourceController.execute(ResourceController.java:88) at hudson.model.Executor.run(Executor.java:124) My shell script only write "hello" in a "hello.txt" file on the server, and nothing is done.

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• How do I format a text file for IIS Mailroot Pickup so that it sends an e-mail with attachments?

  - by Ben McCormack

  How do I need to format a text file so that it can be read by an SMTP service to send an e-mail that has an attachment? We have a server where we are using II6 SMTP to send mail from a Pickup folder. The goal is to drop a properly formatted text file into Mailroot\Pickup and then the file will be automatically processed and sent to the correct SMTP recipient. For simple files, this works correctly. Here's an example of a simple file that works (domain names changed): From:[email protected] To:[email protected] Subject:Hello World! Test Body Of The E-mail When I drop a text file containing the above contents into the Mailroot\Pickup folder, it sends correctly. However, I haven't been able to figure out how to get an attachment to work. I found some material that explained how to encode an SMTP attachment and another tool for simple base64 encoding conversion. Using the information on those pages, I came up with the following text: From:[email protected] To:[email protected] Subject:Hello World! MIME-Version: 1.0 Content-Type: text/plain; boundary="Attached" Content-Disposition: inline; --Attached Content-Transfer-Encoding: base64 Content-Type: text/plain; name="attachment.txt" Content-Disposition: attachment; filenamename="attachment.txt" VGhpcyBpcyBhIHRlc3Qgb2Ygc29tZXRoaW5nIHRvIGVuY29kZS4NCk5ldyBsaW5lDQpOZXcgTGlu ZQ0KIkhlbGxvdyIgISEhDQo9PT09ICcgZnNkZnNkZiAxMjM1NDU2MzQzNA== --Attached-- However, when I place the above text in a file and drop it into Mailroot\Pickup, it doesn't send an attachment correctly. Instead, an e-mail shows up with the following in the body of the e-mail: MIME-Version: 1.0 Content-Type: text/plain; boundary="Attached" Content-Disposition: inline; --Attached Content-Transfer-Encoding: base64 Content-Type: text/plain; name="attachment.txt" Content-Disposition: attachment; filenamename="attachment.txt" VGhpcyBpcyBhIHRlc3Qgb2Ygc29tZXRoaW5nIHRvIGVuY29kZS4NCk5ldyBsaW5lDQpOZXcgTGlu ZQ0KIkhlbGxvdyIgISEhDQo9PT09ICcgZnNkZnNkZiAxMjM1NDU2MzQzNA== --Attached-- I can't figure out what I need to do to format the text file so that the SMTP service correctly sends attachments.

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• SFTP, ChrootDirectory and multiple users

  - by mdo

  I need a setup where I can put the contents of several user folders to a DMZ server from where external clients can download it, protocol SFTP, Linux, OpenSSH. To ease administration we want to use one single user for the upload. What does work is to define ChrootDirectory /home/sftp/ in sshd_config, set the according ownership and modes and define a home dir in passwd so that the working directory of the user fits. This is my structure: /home/sftp/uploader/user1/file1.txt /user2/file2.txt The uploader user can write file1.txt and file2.txt to the corresponding folders and by having the user folders (user1, user2) set to the users' primary group + setting SETGUID on the folders the users are able to even delete the files (which is necessary). Only problem: because /home/sftp/ is the chroot base dir the users can change updir and see other users' folders, though not being able to change into because of access rights. Requirement: We want to prevent users to change to /home/sftp/uploader/ and see other users' folders. My requirements are to use SFTP, have one upload user and every user must have write access to his home dir. Obviously it's not an option to use something like ChrootDirectory %h because every path component of the chroot path needs to have limited access rights, so as far as I understand this does not work.

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• Comparing 2 (or 3 Files If Possible) "Line By Line"

  - by PythEch

  I want to find out the differences of 2 (or 3 files if possible) line by line. Diff utils can do this, however it gives inaccurate results. Because, 2 files have exact number of lines which is "134". But diff gives me "Added Lines" and "Removed Lines". However this is wrong, they have exact the same number of lines, there is no added or removed lines. The text files which I want to find differences of them, have only numbers written, maybe that's why that algortihm fails. I couldn't find any option to prevent that, however I may be wrong, I mean there should be an option for that, but again, I couldn't find. This is what I get (5am.txt vs 6am.txt, there is a huge problem): This is what I want (6am.txt vs 7am.txt, still has problems): But, first the first image still has this problem, at the last lines. Edit: After I figured out that there is no utility to do this, I handled myself. I almost did the same thing as what RedGrittyBrick have done. This script imitates diff utility so I (or you) can use it with diff2html. To use it with diff2html, just change line diff_stdout = os.popen("diff %s" % string.join(argv[1:]), "r") to diff_stdout = os.popen("script.py %s" % string.join(argv[1:]), "r") and name this script whatever you want: import sys f1=open(sys.argv[1],"r") f1_read=f1.readlines() f1.close() f2=open(sys.argv[2],"r") f2_read=f2.readlines() f2.close() changed={} first_c = "" for n in range(len(f1_read)): if f1_read[n]!=f2_read[n]: if first_c == "": first_c=n+1 changed[first_c]=n+1 else: first_c="" #Let's imitate diff-utils... for (x, y) in changed.items(): print "%d,%dc%d,%d" % (x,y,x,y) for i in range(x,y+1): sys.stdout.write("< %s" % f1_read[i-1]) print "---" for i in range(x,y+1): sys.stdout.write("> %s" % f2_read[i-1]) Final results:

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• recursively "normalize" filenames

  - by user66732

  i have made a script, that can recursively rename files to get rid of special chars, etc. in filenames e.g.: before: THIS i.s my file (1).txt after running the script: This-i-s-my-file-1.txt Ok. here it is: But: when i wanted to test it "fully", with filenames like this: ¤¥¦§¨©ª«¬®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÂÃÄÅÆÇÈÊËÌÎÏÐÑÒÔÕ×ØÙUÛUÝÞßàâãäåæçèêëìîïðñòôõ÷øùûýþÿ.txt áíüuúöoóéÁÍÜUÚÖOÓÉ!"#$%&'()*+,:;<=>?@[\]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£.txt it fails: $ sh renamer.sh directorythathasthefiles mv: cannot stat `./áíüuúöoóéÁÍÜUÚÖOÓÉ!"#$%&\'()*+,:;<=>?@[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory mv: cannot stat `./áíüuúöoóéÁÍÜUÚÖOÓÉ!"#$%&\'()*+,:;<=>?@[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory mv: cannot stat `./áíüuúöoóéÁÍÜUÚÖOÓÉ!"#$%&\'()*+,:;<=>?@[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory mv: cannot stat `./áíüuúöoóéÁÍÜUÚÖOÓÉ!"#$%&\'()*+,:;<=>?@[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory mv: cannot stat `./áíüuúöoóéÁÍÜUÚÖOÓÉ!"#$%&\'()*+,:;<=>?@[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory mv: cannot stat `./áíüuúöoóéÁÍÜUÚÖOÓÉ!"#$%&\'()*+,:;<=>?@[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory ...and so on so "mv" can't handle special chars.. :\ i worked on it for many hours.. does anyone has a working one? [that can handle chars [filenames] in that 2 lines too?] Q on pastebin: http://pastebin.com/raw.php?i=19iYZpwY

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• recursively "normalize" filenames

  - by user62367

  i mean getting rid of special chars in filenames, etc. i have made a script, that can recursively rename files [http://pastebin.com/raw.php?i=kXeHbDQw]: e.g.: before: THIS i.s my file (1).txt after running the script: This-i-s-my-file-1.txt Ok. here it is: But: when i wanted to test it "fully", with filenames like this: ¤¥¦§¨©ª«¬®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÂÃÄÅÆÇÈÊËÌÎÏÐÑÒÔÕ×ØÙUÛUÝÞßàâãäåæçèêëìîïðñòôõ÷øùûýþÿ.txt áíüuúöoóéÁÍÜUÚÖOÓÉ!"#$%&'()*+,:;<=>?@[\]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£.txt it fails [http://pastebin.com/raw.php?i=iu8Pwrnr]: $ sh renamer.sh directorythathasthefiles mv: cannot stat `./áíüuúöoóéÁÍÜUÚÖOÓÉ!"#$%&\'()*+,:;<=>?@[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory mv: cannot stat `./áíüuúöoóéÁÍÜUÚÖOÓÉ!"#$%&\'()*+,:;<=>?@[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory mv: cannot stat `./áíüuúöoóéÁÍÜUÚÖOÓÉ!"#$%&\'()*+,:;<=>?@[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory mv: cannot stat `./áíüuúöoóéÁÍÜUÚÖOÓÉ!"#$%&\'()*+,:;<=>?@[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory mv: cannot stat `./áíüuúöoóéÁÍÜUÚÖOÓÉ!"#$%&\'()*+,:;<=>?@[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory mv: cannot stat `./áíüuúöoóéÁÍÜUÚÖOÓÉ!"#$%&\'()*+,:;<=>?@[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory mv: cannot stat `./áíüuúöoóéÁÍÜUÚÖOÓÉ!"#$%&\'()*+,:;<=>?@[]^_`{|}~€‚ƒ„…†....and so on $ so "mv" can't handle special chars.. :\ i worked on it for many hours.. does anyone has a working one? [that can handle chars [filenames] in that 2 lines too?]

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• Linux: Can't overwrite files on samba store

  - by jonescb

  I'm using CentOS 5.5 with smbclient 3.0.33-3.28-el5 (latest version in repo), and I can't overwrite files in my Samba store. I am not the admin for the Samba server, so there isn't anything I can do server side. But I do have write permission to the server. I know the server runs Windows XP or Server 2003; I don't know. I can delete the file, and then copy the new version over, but I can't overwrite it. Using the cp command I'll get this error: [jonescb@localhost ~]$ cp foo.txt /mnt/si_storage/foo.txt cp: cannot create regular file `/mnt/si_storage/foo.txt': No such file or directory` And if I edit a file on the server using vim, I can save it once, but if I save it again I get this: "/mnt/si_storage/foo.txt" E212: Can't open file for writing This is my /etc/fstab entry for the samba server: //192.168.1.2/SI_STORAGE /mnt/si_storage cifs username=myuser,password=mypass 0 0 Edit: I can overwrite files just fine on my XP machine. The CentOS box is the only one having problems.

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• Computing Number of Bits in Public Key

  - by eb80

  I am working with DKIM and trying to compute the public key size of some DKIM signatures. I know from tools that Gmail's is now 2048, but how could I have figured this out myself (i.e., what exact Linux commands and why)? user@host$ dig txt 20120113._domainkey.gmail.com ; <<>> DiG 9.8.3-P1 <<>> txt 20120113._domainkey.gmail.com ;; global options: +cmd ;; Got answer: ;; ->>HEADER<<- opcode: QUERY, status: NOERROR, id: 52228 ;; flags: qr rd ra; QUERY: 1, ANSWER: 1, AUTHORITY: 0, ADDITIONAL: 0 ;; QUESTION SECTION: ;20120113._domainkey.gmail.com. IN TXT ;; ANSWER SECTION: 20120113._domainkey.gmail.com. 300 IN TXT "k=rsa\; p=MIIBIjANBgkqhkiG9w0BAQEFAAOCAQ8AMIIBCgKCAQEA1Kd87/UeJjenpabgbFwh+eBCsSTrqmwIYYvywlbhbqoo2DymndFkbjOVIPIldNs/m40KF+yzMn1skyoxcTUGCQs8g3FgD2Ap3ZB5DekAo5wMmk4wimDO+U8QzI3SD0" "7y2+07wlNWwIt8svnxgdxGkVbbhzY8i+RQ9DpSVpPbF7ykQxtKXkv/ahW3KjViiAH+ghvvIhkx4xYSIc9oSwVmAl5OctMEeWUwg8Istjqz8BZeTWbf41fbNhte7Y+YqZOwq1Sd0DbvYAD9NOZK9vlfuac0598HY+vtSBczUiKERHv1yRbcaQtZFh5wtiRrN04BLUTD21MycBX5jYchHjPY/wIDAQAB" ;; Query time: 262 msec ;; SERVER: 8.8.8.8#53(8.8.8.8) ;; WHEN: Mon Nov 19 10:52:06 2012 ;; MSG SIZE rcvd: 462

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• I need to take the first three letters of a filename and set it into a text file, in a certain fashion. How can I do this?

  - by JuniorD

  Okay, so I need to take the first three letters of a file from a list of files, and place this into a text file in a certain manner. I will provide examples below. Lets say that I have two file names in the same directory, one called cougar.txt and the other bear.txt. These are in the animals directory. I need to take the first three letters of these words, and transpose them into a text file along with the directory, in the following format: BEA = "animals/bear.txt" COU = "animals/cougar.txt" This should happen with any random thing that might be in the list. I'm fairly new to this sort of coding, so I'm not quite sure which language to use, and I'm learning as I go. This new challenge seems fairly daunting to me, and I would be much appreciated if you guys could help. Also, I'm using Windows 7. Been attempting at this all day, to no avail. Preferably done in batch, but if that is impossible I'm open to recommendations.

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• Need some help on how to replay the last game of a java maze game

  - by Marty

  Hello, I am working on creating a Java maze game for a project. The maze is displayed on the console as standard output not in an applet. I have created most of hte code I need, however I am stuck at one problem and that is I need a user to be able to replay the last game i.e redraw the maze with the users moves but without any input from the user. I am not sure on what course of action to take, i was thinking about copying each users move or the position of each move into another array, as you can see i have 2 variables which hold the position of the player, plyrX and plyrY do you think copying these values into a new array after each move would solve my problem and how would i go about this? I have updated my code, apologies about the textIO.java class not being present, not sure how to resolve that exept post a link to TextIO.java [TextIO.java][1] My code below is updated with a new array of type char to hold values from the original maze (read in from text file and displayed using unicode characters) and also to new variables c_plyrX and c_plyrY which I am thinking should hold the values of plyrX and plyrY and copy them into the new array. When I try to call the replayGame(); method from the menu the maze loads for a second then the console exits so im not sure what I am doing wrong Thanks public class MazeGame { //unicode characters that will define the maze walls, //pathways, and in game characters. final static char WALL = '\u2588'; //wall final static char PATH = '\u2591'; //pathway final static char PLAYER = '\u25EF'; //player final static char ENTRANCE = 'E'; //entrance final static char EXIT = '\u2716'; //exit //declaring member variables which will hold the maze co-ordinates //X = rows, Y = columns static int entX = 0; //entrance X co-ordinate static int entY = 1; //entrance y co-ordinate static int plyrX = 0; static int plyrY = 1; static int exitX = 24; //exit X co-ordinate static int exitY = 37; //exit Y co-ordinate //static member variables which hold maze values //used so values can be accessed from different methods static int rows; //rows variable static int cols; //columns variable static char[][] maze; //defines 2 dimensional array to hold the maze //variables that hold player movement values static char dir; //direction static int spaces; //amount of spaces user can travel //variable to hold amount of moves the user has taken; static int movesTaken = 0; //new array to hold player moves for replaying game static char[][] mazeCopy; static int c_plyrX; static int c_plyrY; /** userMenu method for displaying the user menu which will provide various options for * the user to choose such as play a maze game, get instructions, etc. */ public static void userMenu(){ TextIO.putln("Maze Game"); TextIO.putln("*********"); TextIO.putln("Choose an option."); TextIO.putln(""); TextIO.putln("1. Play the Maze Game."); TextIO.putln("2. View Instructions."); TextIO.putln("3. Replay the last game."); TextIO.putln("4. Exit the Maze Game."); TextIO.putln(""); int option; //variable for holding users option TextIO.put("Type your choice: "); option = TextIO.getlnInt(); //gets users option //switch statement for processing menu options switch(option){ case 1: playMazeGame(); case 2: instructions(); case 3: if (c_plyrX == plyrX && c_plyrY == plyrY)replayGame(); else { TextIO.putln("Option not available yet, you need to play a game first."); TextIO.putln(); userMenu(); } case 4: System.exit(0); //exits the user out of the console default: TextIO.put("Option must be 1, 2, 3 or 4"); } } //end of userMenu /**main method, will call the userMenu and get the users choice and call * the relevant method to execute the users choice. */ public static void main(String[]args){ userMenu(); //calls the userMenu method } //end of main method /**instructions method, displays instructions on how to play * the game to the user/ */ public static void instructions(){ TextIO.putln("To beat the Maze Game you have to move your character"); TextIO.putln("through the maze and reach the exit in as few moves as possible."); TextIO.putln(""); TextIO.putln("Your characer is displayed as a " + PLAYER); TextIO.putln("The maze exit is displayed as a " + EXIT); TextIO.putln("Reach the exit and you have won escaped the maze."); TextIO.putln("To control your character type the direction you want to go"); TextIO.putln("and how many spaces you want to move"); TextIO.putln("for example 'D3' will move your character"); TextIO.putln("down 3 spaces."); TextIO.putln("Remember you can't walk through walls!"); boolean insOption; //boolean variable TextIO.putln(""); TextIO.put("Do you want to play the Maze Game now? (Y or N) "); insOption = TextIO.getlnBoolean(); if (insOption == true)playMazeGame(); else userMenu(); } //end of instructions method /**playMazeGame method, calls the loadMaze method and the charMove method * to start playing the Maze Game. */ public static void playMazeGame(){ loadMaze(); plyrMoves(); } //end of playMazeGame method /**loadMaze method, loads the 39x25 maze from the MazeGame.txt text file * and inserts values from the text file into the maze array and * displays the maze on screen using the unicode block characters. * plyrX and plyrY variables are set at their staring co ordinates so that when * a game is completed and the user selects to play a new game * the player character will always be at position 01. */ public static void loadMaze(){ plyrX = 0; plyrY = 1; TextIO.readFile("MazeGame.txt"); //now reads from the external MazeGame.txt file rows = TextIO.getInt(); //gets the number of rows from text file to create X dimensions cols = TextIO.getlnInt(); //gets number of columns from text file to create Y dimensions maze = new char[rows][cols]; //creates maze array of base type char with specified dimnensions //loop to process the array and read in values from the text file. for (int i = 0; i<rows; i++){ for (int j = 0; j<cols; j++){ maze[i][j] = TextIO.getChar(); } TextIO.getln(); } //end for loop TextIO.readStandardInput(); //closes MazeGame.txt file and reads from //standard input. //loop to process the array values and display as unicode characters for (int i = 0; i<rows; i++){ for (int j = 0; j<cols; j++){ if (i == plyrX && j == plyrY){ plyrX = i; plyrY = j; TextIO.put(PLAYER); //puts the player character at player co-ords } else{ if (maze[i][j] == '0') TextIO.putf("%c",WALL); //puts wall block if (maze[i][j] == '1') TextIO.putf("%c",PATH); //puts path block if (maze[i][j] == '2') { entX = i; entY = j; TextIO.putf("%c",ENTRANCE); //puts entrance character } if (maze[i][j] == '3') { exitX = i; //holds value of exit exitY = j; //co-ordinates TextIO.putf("%c",EXIT); //puts exit character } } } TextIO.putln(); } //end for loop } //end of loadMaze method /**redrawMaze method, method for redrawing the maze after each move. * */ public static void redrawMaze(){ TextIO.readFile("MazeGame.txt"); //now reads from the external MazeGame.txt file rows = TextIO.getInt(); //gets the number of rows from text file to create X dimensions cols = TextIO.getlnInt(); //gets number of columns from text file to create Y dimensions maze = new char[rows][cols]; //creates maze array of base type char with specified dimnensions //loop to process the array and read in values from the text file. for (int i = 0; i<rows; i++){ for (int j = 0; j<cols; j++){ maze[i][j] = TextIO.getChar(); } TextIO.getln(); } //end for loop TextIO.readStandardInput(); //closes MazeGame.txt file and reads from //standard input. //loop to process the array values and display as unicode characters for (int i = 0; i<rows; i++){ for (int j = 0; j<cols; j++){ if (i == plyrX && j == plyrY){ plyrX = i; plyrY = j; TextIO.put(PLAYER); //puts the player character at player co-ords } else{ if (maze[i][j] == '0') TextIO.putf("%c",WALL); //puts wall block if (maze[i][j] == '1') TextIO.putf("%c",PATH); //puts path block if (maze[i][j] == '2') { entX = i; entY = j; TextIO.putf("%c",ENTRANCE); //puts entrance character } if (maze[i][j] == '3') { exitX = i; //holds value of exit exitY = j; //co-ordinates TextIO.putf("%c",EXIT); //puts exit character } } } TextIO.putln(); } //end for loop } //end redrawMaze method /**replay game method * */ public static void replayGame(){ c_plyrX = plyrX; c_plyrY = plyrY; TextIO.readFile("MazeGame.txt"); //now reads from the external MazeGame.txt file rows = TextIO.getInt(); //gets the number of rows from text file to create X dimensions cols = TextIO.getlnInt(); //gets number of columns from text file to create Y dimensions mazeCopy = new char[rows][cols]; //creates maze array of base type char with specified dimnensions //loop to process the array and read in values from the text file. for (int i = 0; i<rows; i++){ for (int j = 0; j<cols; j++){ mazeCopy[i][j] = TextIO.getChar(); } TextIO.getln(); } //end for loop TextIO.readStandardInput(); //closes MazeGame.txt file and reads from //standard input. //loop to process the array values and display as unicode characters for (int i = 0; i<rows; i++){ for (int j = 0; j<cols; j++){ if (i == c_plyrX && j == c_plyrY){ c_plyrX = i; c_plyrY = j; TextIO.put(PLAYER); //puts the player character at player co-ords } else{ if (mazeCopy[i][j] == '0') TextIO.putf("%c",WALL); //puts wall block if (mazeCopy[i][j] == '1') TextIO.putf("%c",PATH); //puts path block if (mazeCopy[i][j] == '2') { entX = i; entY = j; TextIO.putf("%c",ENTRANCE); //puts entrance character } if (mazeCopy[i][j] == '3') { exitX = i; //holds value of exit exitY = j; //co-ordinates TextIO.putf("%c",EXIT); //puts exit character } } } TextIO.putln(); } //end for loop } //end replayGame method /**plyrMoves method, method for moving the players character * around the maze. */ public static void plyrMoves(){ int nplyrX = plyrX; int nplyrY = plyrY; int pMoves; direction(); //UP if (dir == 'U' || dir == 'u'){ nplyrX = plyrX; nplyrY = plyrY; for(pMoves = 0; pMoves <= spaces; pMoves++){ if (maze[nplyrX][nplyrY] == '0'){ TextIO.putln("Invalid move, try again."); } else if (pMoves != spaces){ nplyrX =plyrX + 1; } else { plyrX = plyrX-spaces; c_plyrX = plyrX; movesTaken++; } } }//end UP if //DOWN if (dir == 'D' || dir == 'd'){ nplyrX = plyrX; nplyrY = plyrY; for (pMoves = 0; pMoves <= spaces; pMoves ++){ if (maze[nplyrX][nplyrY] == '0'){ TextIO.putln("Invalid move, try again"); } else if (pMoves != spaces){ nplyrX = plyrX+1; } else{ plyrX = plyrX+spaces; c_plyrX = plyrX; movesTaken++; } } } //end DOWN if //LEFT if (dir == 'L' || dir =='l'){ nplyrX = plyrX; nplyrY = plyrY; for (pMoves = 0; pMoves <= spaces; pMoves++){ if (maze[nplyrX][nplyrY] == '0'){ TextIO.putln("Invalid move, try again"); } else if (pMoves != spaces){ nplyrY = plyrY + 1; } else{ plyrY = plyrY-spaces; c_plyrY = plyrY; movesTaken++; } } } //end LEFT if //RIGHT if (dir == 'R' || dir == 'r'){ nplyrX = plyrX; nplyrY = plyrY; for (pMoves = 0; pMoves <= spaces; pMoves++){ if (maze[nplyrX][nplyrY] == '0'){ TextIO.putln("Invalid move, try again."); } else if (pMoves != spaces){ nplyrY += 1; } else{ plyrY = plyrY+spaces; c_plyrY = plyrY; movesTaken++; } } } //end RIGHT if //prints message if player escapes from the maze. if (maze[plyrX][plyrY] == '3'){ TextIO.putln("****Congratulations****"); TextIO.putln(); TextIO.putln("You have escaped from the maze."); TextIO.putln(); userMenu(); } else{ movesTaken++; redrawMaze(); plyrMoves(); } } //end of plyrMoves method /**direction, method * */ public static char direction(){ TextIO.putln("Enter the direction you wish to move in and the distance"); TextIO.putln("i.e D3 = move down 3 spaces"); TextIO.putln("U - Up, D - Down, L - Left, R - Right: "); dir = TextIO.getChar(); if (dir =='U' || dir == 'D' || dir == 'L' || dir == 'R' || dir == 'u' || dir == 'd' || dir == 'l' || dir == 'r'){ spacesMoved(); } else{ loadMaze(); TextIO.putln("Invalid direction!"); TextIO.put("Direction must be one of U, D, L or R"); direction(); } return dir; //returns the value of dir (direction) } //end direction method /**spaces method, gets the amount of spaces the user wants to move * */ public static int spacesMoved(){ TextIO.putln(" "); spaces = TextIO.getlnInt(); if (spaces <= 0){ loadMaze(); TextIO.put("Invalid amount of spaces, try again"); spacesMoved(); } return spaces; } //end spacesMoved method } //end of MazeGame class

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• NTFS Issues in Windows 7 and 2008 R2 - 'Is it a Bug?'

  - by renewieldraaijer

  I have been using the various versions of the Microsoft Windows product line since NT4 and I really thought I knew the ins and outs about the NTFS filesystem by now. There were always a few rules of thumb to understand what happens if you move data around. These rules were: "If you copy data, the copied data will inherit the permissions of the location it is being copied to. The same goes for moving data between disk partitions. Only when you move data within the same partition, the permissions are kept."  Recently I was asked to assist in troubleshooting some NTFS related issues. This forced me to have another good look at this theory. To my surprise I found out that this theory does not completely stand anymore. Apparently some things have changed since the release of Windows Vista / Windows 2008. Since the release of these Operating Systems, a move within the same disk partition results in the data inheriting the permissions of the location it is being copied into. A major change in the NTFS filesystem you would think!  Not quite! The above only counts when the move operation is being performed by using Windows Explorer. A move by using the 'move' command from within a cmd prompt for example, retains the NTFS permissions, just like before in Windows XP and older systems. Conclusion: The Windows Explorer is responsible for changing the ACL's of the moved data. This is a remarkable change, but if you follow this theory, the resulting ACL after a move operation is still predictable.  We could say that since Windows Vista and Windows 2008, a new rule set applies: "If you copy data, the copied data will inherit the permissions of the location it is being copied to. Same goes for moving data between disk partitions and within disk partitions. Only when you move data within the same partition by using something else than the Windows Explorer, the permissions are kept." The above behavior should be unchanged in Windows 7 / Windows 2008 R2, compared to Windows Vista / 2008. But somehow the NTFS permissions are not so predictable in Windows 7 and Windows 2008 R2. Moving data within the same disk partition the one time results in the permissions being kept and the next time results in inherited permissions from the destination location. I will try to demonstrate this in a few examples: Example 1 (Incorrect behavior): Consider two folders, 'Folder A' and 'Folder B' with the following permissions configured.                    Now we create the test file 'test file 1.txt' in 'Folder A' and afterwards move this file to 'Folder B' using Windows Explorer.                       According to the new theory, the file should inherit the permissions of 'Folder B' and therefore 'Group B' should appear in the ACL of 'test file 1.txt'. In the screenshot below the resulting permissions are displayed. The permissions from the originating location are kept, while the permissions of 'Folder B' should be inherited.                   Example 2 (Correct behavior): Again, consider the same two folders. This time we make a small modification to the ACL of 'Folder A'. We add 'Group C' to the ACL and again we create a file in 'Folder A' which we name 'test file 2.txt'.                    Next, we move 'test file 2.txt' to 'Folder B'.                       Again, we check the permissions of 'test file 2.txt' at the target location. We can now see that the permissions are inherited. This is what should be happening, and can be considered 'correct behavior' for Windows Vista / 2008 / 7 / 2008 R2. It remains uncertain why this behavior is so inconsistent. At this time, this is under investigation with Microsoft Support. The investigation has been going for the last two weeks and it is beginning to look like there is no rational reason for this, other than a bug in the Windows Explorer in Windows 7 and 2008 R2. As soon as there is any certainty on this, I will note it here in this blog.                   The examples above are harmless tests, by using my own laptop. If you would create the same set of folders and groups, and configure exactly the same permissions, you will see exactly the same behavior. Be sure to use Windows 7 or Windows 2008 R2.   Initially the problem arose at a customer site where move operations on data on the fileserver by users would result in unpredictable results. This resulted in the wrong set of people having àccess permissions on data that they should not have permissions to. Off course this is something we want to prevent at all costs.   I have also done several tests with move operations by using the move command in a cmd prompt. This way the behavior is always consistent. The inconsistent behavior is only exposed when using the Windows Explorer to initiate the move operation, and only when using Windows 7 or Windows 2008 R2 systems. It is evident that this behavior changes when the ACL of a folder has been changed, for example by adding an extra entry. The reason for this remains uncertain though. To be continued…. A dutch version of this post can be found at: http://blogs.platani.nl/?p=612

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• error in finding out the lexems and no of lines of a text file in C

  - by mekasperasky

  #include<stdio.h> #include<ctype.h> #include<string.h> int main() { int i=0,j,k,lines_count[2]={1,1},operand_count[2]={0},operator_count[2]={0},uoperator_count[2]={0},control_count[2]={0,0},cl[13]={0},variable_dec[2]={0,0},l,p[2]={0},ct,variable_used[2]={0,0},constant_count[2],s[2]={0},t[2]={0}; char a,b[100],c[100]; char d[100]={0}; j=30; FILE *fp1[2],*fp2; fp1[0]=fopen("program1.txt","r"); fp1[1]=fopen("program2.txt","r"); //the source file is opened in read only mode which will passed through the lexer fp2=fopen("ccv1ouput.txt","wb"); //now lets remove all the white spaces and store the rest of the words in a file if(fp1[0]==NULL) { perror("failed to open program1.txt"); //return EXIT_FAILURE; } if(fp1[1]==NULL) { perror("failed to open program2.txt"); //return EXIT_FAILURE; } i=0; k=0; ct=0; while(ct!=2) { while(!feof(fp1[ct])) { a=fgetc(fp1[ct]); if(a!=' '&&a!='\n') { if (!isalpha(a) && !isdigit(a)) { switch(a) { case '+':{ i=0; cl[0]=1; operator_count[ct]=operator_count[ct]+1;break;} case '-':{ cl[1]=1; operator_count[ct]=operator_count[ct]+1;i=0;break;} case '*':{ cl[2]=1; operator_count[ct]=operator_count[ct]+1;i=0;break;} case '/':{ cl[3]=1; operator_count[ct]=operator_count[ct]+1;i=0;break;} case '=':{a=fgetc(fp1[ct]); if (a=='='){cl[4]=1; operator_count[ct]=operator_count[ct]+1; operand_count[ct]=operand_count[ct]+1;} else { cl[5]=1; operator_count[ct]=operator_count[ct]+1; operand_count[ct]=operand_count[ct]+1; ungetc(1,fp1[ct]); } break;} case '%':{ cl[6]=1; operator_count[ct]=operator_count[ct]+1;i=0;break;} case '<':{ a=fgetc(fp1[ct]); if (a=='=') {cl[7]=1; operator_count[ct]=operator_count[ct]+1;} else { cl[8]=1; operator_count[ct]=operator_count[ct]+1; ungetc(1,fp1[ct]); } break; } case '>':{ ; a=fgetc(fp1[ct]); if (a=='='){cl[9]=1; operator_count[ct]=operator_count[ct]+1;} else { cl[10]=1; operator_count[ct]=operator_count[ct]+1; ungetc(1,fp1[ct]); } break;} case '&':{ cl[11]=1; a=fgetc(fp1[ct]); operator_count[ct]=operator_count[ct]+1; operand_count[ct]=operand_count[ct]+1; variable_used[ct]=variable_used[ct]-1; break; } case '|':{ cl[12]=1; a=fgetc(fp1[ct]); operator_count[ct]=operator_count[ct]+1; operand_count[ct]=operand_count[ct]+1; variable_used[ct]=variable_used[ct]-1; break; } case '#':{ while(a!='\n') { a=fgetc(fp1[ct]); } } } } else { d[i]=a; i=i+1; k=k+1; } } else { //printf("%s \n",d); if((strcmp(d,"if")==0)){ memset ( d, 0, 100 ); i=0; control_count[ct]=control_count[ct]+1; } else if(strcmp(d,"then")==0){ i=0;memset ( d, 0, 100 );control_count[ct]=control_count[ct]+1;} else if(strcmp(d,"else")==0){ i=0;memset ( d, 0, 100 );control_count[ct]=control_count[ct]+1;} else if(strcmp(d,"while")==0){ i=0;memset ( d, 0, 100 );control_count[ct]=control_count[ct]+1;} else if(strcmp(d,"int")==0){ while(a != '\n') { a=fgetc(fp1[ct]); if (isalpha(a) ) variable_dec[ct]=variable_dec[ct]+1; } memset ( d, 0, 100 ); lines_count[ct]=lines_count[ct]+1; } else if(strcmp(d,"char")==0){while(a != '\n') { a=fgetc(fp1[ct]); if (isalpha(a) ) variable_dec[ct]=variable_dec[ct]+1; } memset ( d, 0, 100 ); lines_count[ct]=lines_count[ct]+1; } else if(strcmp(d,"float")==0){while(a != '\n') { a=fgetc(fp1[ct]); if (isalpha(a) ) variable_dec[ct]=variable_dec[ct]+1; } memset ( d, 0, 100 ); lines_count[ct]=lines_count[ct]+1; } else if(strcmp(d,"printf")==0){while(a!='\n') a=fgetc(fp1[ct]); memset(d,0,100); } else if(strcmp(d,"scanf")==0){while(a!='\n') a=fgetc(fp1[ct]); memset(d,0,100);} else if (isdigit(d[i-1])) { memset ( d, 0, 100 ); i=0; constant_count[ct]=constant_count[ct]+1; operand_count[ct]=operand_count[ct]+1; } else if (isalpha(d[i-1]) && strcmp(d,"int")!=0 && strcmp(d,"char")!=0 && strcmp(d,"float")!=0 && (strcmp(d,"if")!=0) && strcmp(d,"then")!=0 && strcmp(d,"else")!=0 && strcmp(d,"while")!=0 && strcmp(d,"printf")!=0 && strcmp(d,"scanf")!=0) { memset ( d, 0, 100 ); i=0; operand_count[ct]=operand_count[ct]+1; } else if(a=='\n') { lines_count[ct]=lines_count[ct]+1; memset ( d, 0, 100 ); } } } fclose(fp1[ct]); operand_count[ct]=operand_count[ct]-5; variable_used[0]=operand_count[0]-constant_count[0]; variable_used[1]=operand_count[1]-constant_count[1]; for(j=0;j<12;j++) uoperator_count[ct]=uoperator_count[ct]+cl[j]; fprintf(fp2,"\n statistics of program %d",ct+1); fprintf(fp2,"\n the no of lines ---> %d",lines_count[ct]); fprintf(fp2,"\n the no of operands --->%d",operand_count[ct]); fprintf(fp2,"\n the no of operator --->%d",operator_count[ct]); fprintf(fp2,"\n the no of control statments --->%d",control_count[ct]); fprintf(fp2,"\n the no of unique operators --->%d",uoperator_count[ct]); fprintf(fp2,"\n the no of variables declared--->%d",variable_dec[ct]); fprintf(fp2,"\n the no of variables used--->%d",variable_used[ct]); fprintf(fp2,"\n ---------------------------------"); fprintf(fp2,"\n \t \t \t"); ct=ct+1; } t[0]=lines_count[0]+control_count[0]+uoperator_count[0]; t[1]=lines_count[1]+control_count[1]+uoperator_count[1]; s[0]=operator_count[0]+operand_count[0]+variable_dec[0]+variable_used[0]; s[1]=operator_count[1]+operand_count[1]+variable_dec[1]+variable_used[1]; fprintf(fp2,"\n the time complexity of program 1 is %d",t[0]); fprintf(fp2,"\n the time complexity of program 2 is %d",t[1]); fprintf(fp2,"\n the space complexity of program 1 is %d",s[0]); fprintf(fp2,"\n the space complexity of program 2 is %d",s[1]); if((t[0]>t[1]) && (s[0] >s[1])) fprintf(fp2,"\n the efficiency of program 2 is greater than program 1"); else if(t[0]<t[1] && s[0] < s[1]) fprintf(fp2,"\n the efficiency of program 1 is greater than program 2 " ); else if (t[0]+s[0] > t[1]+s[1]) fprintf(fp2,"\n the efficiency of program 1 is greater than program 2"); else if (t[0]+s[0] < t[1]+s[1]) fprintf(fp2,"\n the efficiency of program 2 is greater than program 1"); else if (t[0]+s[0] == t[1]+s[1]) fprintf(fp2,"\n the efficiency of program 1 is equal to that of program 2"); fclose(fp2); return 0; } this code basically compares two c codes and finds out the no. of variables declared , used , no. of control statements , no. of lines and no. of unique operators , and operands , so as to find out the time complexity and space complexity of of the two programs given in the text file program1.txt and program2.txt ... Lets say program1.txt is this #include<stdio.h> #include<math.h> int main () { FILE *fp; fp=fopen("output.txt","w"); long double t,y=0,x=0,e=5,f=1,w=1; for (t=0;t<10;t=t+0.01) { //if (isnan(y) || isinf(y)) //break; fprintf(fp,"%ld\t%ld\n",y,x); y = y + ((e*(1 - (x*x))*y) - x + f*cos(w*0.1))*0.1; x = x + y*0.1; } fclose(fp); return (0); } i havent indented it as its just a text file . But my output is totally faulty . Its not able to find the any of the ouput that i need . Where is the bug in this ? I am not able to figure out as the algorithm looks fine .

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• javaf, problem...plz help someone...urgent [closed]

  - by innovative_aj

  i have made a word guessing game, when i click myButton to check if the guessed word is right or wrong, ball1 is moved into the "container" if its right, i want that when i click the button again and if the typed word is right, the 2nd ball should move into the container too... means one ball per correct answer...plz help me someone and provide me with the code that i can implement, its quite urgent... controller class coding /* * To change this template, choose Tools | Templates * and open the template in the editor. */ package project3; import java.net.URL; import java.util.ResourceBundle; import javafx.event.ActionEvent; import javafx.event.EventHandler; import javafx.fxml.FXML; import javafx.fxml.Initializable; import javafx.scene.control.Button; import javafx.scene.control.Label; import javafx.scene.control.TextField; import javafx.scene.layout.StackPane; import javafx.scene.shape.Circle; /** * FXML Controller class * * @xxx */ public class MyFxmlController implements Initializable { @FXML // fx:id="ball1" private Circle ball1; // Value injected by FXMLLoader @FXML // fx:id="ball2" private Circle ball2; // Value injected by FXMLLoader @FXML // fx:id="ball3" private Circle ball3; // Value injected by FXMLLoader @FXML // fx:id="ball4" private Circle ball4; // Value injected by FXMLLoader @FXML // fx:id="container" private Circle container; // Value injected by FXMLLoader @FXML // fx:id="myButton" private Button myButton; // Value injected by FXMLLoader @FXML // fx:id="myLabel1" private Label myLabel1; // Value injected by FXMLLoader @FXML // fx:id="myLabel2" private Label myLabel2; // Value injected by FXMLLoader @FXML // fx:id="pane" private StackPane pane; // Value injected by FXMLLoader @FXML // fx:id="txt" private TextField txt; // Value injected by FXMLLoader @Override // This method is called by the FXMLLoader when initialization is complete public void initialize(URL fxmlFileLocation, ResourceBundle resources) { assert ball1 != null : "fx:id=\"ball1\" was not injected: check your FXML file 'MyFxml.fxml'."; assert ball2 != null : "fx:id=\"ball2\" was not injected: check your FXML file 'MyFxml.fxml'."; assert ball3 != null : "fx:id=\"ball3\" was not injected: check your FXML file 'MyFxml.fxml'."; assert ball4 != null : "fx:id=\"ball4\" was not injected: check your FXML file 'MyFxml.fxml'."; assert container != null : "fx:id=\"container\" was not injected: check your FXML file 'MyFxml.fxml'."; assert myButton != null : "fx:id=\"myButton\" was not injected: check your FXML file 'MyFxml.fxml'."; assert myLabel1 != null : "fx:id=\"myLabel1\" was not injected: check your FXML file 'MyFxml.fxml'."; assert myLabel2 != null : "fx:id=\"myLabel2\" was not injected: check your FXML file 'MyFxml.fxml'."; assert pane != null : "fx:id=\"pane\" was not injected: check your FXML file 'MyFxml.fxml'."; assert txt != null : "fx:id=\"txt\" was not injected: check your FXML file 'MyFxml.fxml'."; // initialize your logic here: all @FXML variables will have been injected myButton.setOnAction(new EventHandler<ActionEvent>(){ @Override public void handle(ActionEvent event) { int count = 0; String guessed=txt.getText(); boolean result; result=MyCode.check(guessed); if(result) { ball1.setTranslateX(600); ball1.setTranslateY(250-container.getRadius()); //ball2.setTranslateX(600); // ball2.setTranslateY(250-container.getRadius()); } else System.out.println("wrong"); } }); } } word guessing logic public class MyCode { static String x="Netbeans"; static String y[]={"net","beans","neat","beat","bet"}; //static int counter; // public MyCode() { // counter++; //} static boolean check(String guessed) { int count=0; boolean result=false; //counter++; //System.out.println("turns"+counter); for(count=0;count<5;count++) { if(guessed.equals(y[count])) { result=true; break; } } if(result) System.out.println("Right"); else System.out.println("Wrong"); return result; } }

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