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• Linux, how to capture screen, and simulate mouse movements.

  - by 2di

  Hi All I need to capture screen (as print screen) in the way so I can access pixel color data, to do some image recognition, after that I will need to generate mouse events on the screen such as left click, drag and drop (moving mouse while button is pressed, and then release it). Once its done, image will be deleted. Note: I need to capture whole screen everything that user can see, and I need to simulate clicks outside window of my program (if it makes any difference) Spec: Linux ubuntu Language: C++ Performance is not very important,"print screen" function will be executed once every ~10 sec. Duration of the process can be up to 24 hours so method needs to be stable and memory leaks free (as usuall :) I was able to do in windows with win GDI and some windows events, but I'ev no idea how to do it in Linux. Thanks a lot

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• How to manage end user documentation for a project under continuous integration?

  - by mcdon

  I have a project under continuous integration and would like to add end user documentation to the project. The end user documentation is a user manual, not API documentation. In our environment we use windows, c#, msbuild, cruisecontrol.net and subversion. We are currently using DocToHelp to create our help file, which is based on an msword document. I'm looking for some guidance on how to manage the end user documentation. What documentation tools should I use? Should any of the documentation tools be part of the build script? Should the output files from the documentation tool be stored in subversion? What type of help files would be best to use?

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• How can I check if a user has written his username and password correctly?

  - by Sergio Tapia

  I'm using a Linq-to-SQL class called Scans.dbml. In that class I've dragged a table called Users (username, password, role) onto the graphic area and now I can access User object via a UserRepository class: using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace Scanner.Classes { public class UserRepository { private ScansDataContext db = new ScansDataContext(); public User getUser(string username) { return db.Users.SingleOrDefault(x => x.username == username); } public bool exists(string username) { } } } Now in my Login form, I want to use this Linq-to-SQL goodness to do all the data related activities. UserRepository users = new UserRepository(); private void btnLogin_Click(object sender, EventArgs e) { loginToSystem(); } private void loginToSystem() { if (users.getUser(txtUsername.Text)) { } //If txtUsername exists && User.password == Salt(txtPassword) //then Show.MainForm() with User.accountType in constructor to set permissions. } I need help with verifying that a user exists && that that users.password is equal to SALT(txtpassword.text). Any guidance please?

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• Linux - How do i know the block map of the given file and/or the free space map of the partition.

  - by Inso Reiges

  Hello, I am on Linux and need to know either of the two things: 1) If i have a regular file on some file system on a partition under Linux is there a way to know the set of the physical blocks that this file occupies on the drive from user space? Or at least the set of the file system's clusters? 2) Is there a way to get the same information about the whole free space of the given file system? In both cases i understand that if there is any possible way to extract this info it will probably be totally unsafe and racy (anything could happen to these set of blocks between the time i see them and act on them somehow). I also really don't want an implementation that will have to know a lot about every filesystem.

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• How can I isolate the form controls in a ASP Web User Control from the rest of the page's form contr

  - by Justin808

  I have a Web User Control I created for authentication. The web user control is inside the box below. Clicking any button (1 or 2) below works correct as it goes to the correct c# button click event in the code behind file. If I press enter on fields a or b it goes to the correct callback (button1's) if I press enter on field c it still goes to button1's callback, not button2's How can I give my web user control a nice self contained for and view state etc, so it wont mess with the remainder of the page's form? +--------------+ | User: __a___ | | Pass: __b___ | | [button1]| +--------------+ Prompt:______c______ [button2]

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• UX Design Question: Should a multi step wizard save the form contents when the user clicks 'go back'

  - by Ashwin Prabhu

  I am developing a web application that collects data over multiple steps through a wizard. Steps are generally not interdependent, in that data input at each step has little or no effect on the consequent steps. However each step may have a set of validations which determine whether the user can progress to the next step by clicking 'continue' What should be the behavior when the user clicks previous? a Quickly move to the previous page, thus losing all the unsaved data in the form. Prompting the user with a warning is an option, but it can become irritating quite soon. b Move to the previous page saving all the data in the current step - without triggering validations, so that when the user comes back she sees the form in the same state that she left it in. c any other behaviour All opinions are welcome :)

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• How to visually reject user input in a table?

  - by FX

  In the programming of a table-based application module (i.e. the user mostly enters tabular data in an already laid-out table), how would you reject user input for a given cell? The scenario is: the user edits the cell, enters something (text, picture, ...) and you want them to notice when they finish editing (hitting enter, for example) that their entry is not valid for your given "format" (in the wider meaning: it can be that they entered a string instead of a number, that their entry is too long, too short, they include a picture while it's not acceptable, ...). I can see two different things happening: You can rather easily fit their entry into your format, and you do so, but you want them to notice it so they can change if your guess is not good enough (example: they entered "15.47" in a field that needs to be an integer, so your program makes it "15") You cannot guess what to do with their entry, and want to inform them that it's not valid. My question specifically is: what visual display can you offer to inform the user that his input is invalid? Is it preferable to refuse to leave the editing mode, or not? The two things I can imagine are: using colors (red background if invalid, yellow background for my case 1 above) when you reject an input, do something like Apple does for password entry of user accounts: you make the cell "shaking" (i.e. oscillating left and right) for one second, and keep the focus/editing in their so they don't loose what they've typed. Let's hear your suggestions. PS: This question is, at least in my thought process, somehow a continuation and a specialization of my previous question on getting users to read error messages. PPS: Made this community wiki, was that the right thing to do on this kind of question or not?

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• How do I compile on linux to share with all distributions?

  - by Andrew M

  I compiled a PHP extension on Fedora Core 12, but when I send it to someone using CentOS they get the error: "ELF file OS ABI invalid" I'm not sure what causes this running file provides the following info: ELF 64-bit LSB shared object, AMD x86-64, version 1 (GNU/Linux), not stripped An extension that loads fine provides the following from file: ELF 64-bit LSB shared object, AMD x86-64, version 1 (SYSV), not stripped So it seems I need to generate a SYSV type file for some distributions, instead of a GNU/LINUX file, no idea how though. Any pointers? Also should I be statically linking?

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• Linux binary built for 2.0 kernel wouldn't execute on 2.6.x kernel.

  - by lorin

  I was installing a binary Linux application on Ubuntu 9.10 x86_64. The app shipped with an old version of gzip (1.2.4), that was compiled for a much older kernel: $ file gzip gzip: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.0.0, stripped I wasn't able to execute this program. If I tried, this happened: $ ./gzip -bash: ./gzip: No such file or directory ldd was similarly unhappy with this binary: $ ldd gzip not a dynamic executable This isn't a showstopper for me, since my installation has a working version of gzip I can use. But I'm curious: What's the most likely source of this problem? A corrupted file? Or a binary incompatibility due to being built for a much older {kernel,libc,...}?

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• How to set cursor position in C on linux?

  - by frx08

  Hi all, how can I set cursor position through a C program in Linux? thanks :)

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• How do you debug c/c++ source code in linux using emacs?

  - by vmihai

  Hello all, I am using emacs and autotools, to write and compile c/c++ sources on linux. I am using gdb via GUD in emacs. I have defined for convenience: F7:compile, F10:gud-next, F11:gud-step, F5:gud-cont, F9:gud-tbreak, F8:gud-until, F4:gud-print. I am mainly interested in debugging c/c++ source code on linux from emacs and I would like to get the most gdb can give. Unfortunately I am using only F4 which prints the variable under cursor. So my question is how do you guys debug the source code ? What programs do you use ? What key bindings (functionality) do you use mostly ? What do you need the debugger to do for you ? If you do weird stuff it doesn't matter. I would like to know everything to boost my speed a bit here. Thanks in advance. Mihai

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• Having trouble reading XML file from Windows server. Works on Linux

  - by DuFF14

  I'm parsing an XML file in an android app. My success varies depending upon where the file is hosted. After hosting the file on 4 different servers (2 Linux, 2 Windows), I discovered that when the xml is hosted on a Linux server, the app works. When it's hosted on a Windows server, I am unable to parse correctly. Instead of reading the expected xml tags, it reads HTML tags (, , , etc). I'm not sure why it doesn't work on Windows servers, or if that is even the issue and not just a coincidence. Any help is appreciated. Thanks. Here is my code: private void getXmlData() { HttpClient httpclient = new DefaultHttpClient(); String url = XML_URL; HttpPost httppost = new HttpPost(url); HttpResponse response = httpclient.execute(httppost); SaxParser saxParser = new SaxParser(response); parsedXML = saxParser.parse(); }

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• h2 & linux. how to start the database?

  - by David

  sorry im rather new to linux ubuntu. i have an application that i made that runs with tomcat and connects to an h2 database. it all works fine on my windows laptop. im now moving it onto my linux computer to run all the time. but im having trouble with starting the database. i have downloaded h2 from their website. and its sitting on my desktop. the question is. how to start the database. and is their any other commands i need to know to run and operate it. thankyous

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• Do you think its user unfriendly to show error message in tooltips ?

  - by msfanboy

  Hello, when my user enters data validated as wrong a red circle with a white exclamation mark is shown in the right part of the textbox with the wrong data. The error message is only shown when the user hovers the textbox with wrong data. Do you think that is a bad User experience ? I could show the red error message text to the right side of the textboxes if there would still be space...

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• Macros for GCC/G++ to differentiate Linux and Mac OSX?

  - by Viet

  Macros for GCC/G++ to differentiate Linux and Mac OSX?

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• User control event or method override where custom properties are valid?

  - by Curtis White

  I have an ASP.NET user control that is used in another use control. The parent user control uses data-binding to bind to a custom property of the child user control. What method can I override or page event where I am ensured that the property state is set? I think in a page it is PageLoaded versus the Page_Load override? I am looking for this in the user control because my property is always null even though it is set. Thanks.

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• How do I erase printed characters in a console application(Linux)?

  - by Binny V A

  Hi all, I am creating a small console app that needs a progress bar. Something like... Conversion: 175/348 Seconds |========== | 50% My question is, how do you erase characters already printed to the console? When I reach the 51st percentage, I have to erase this line from the console and insert a new line. In my current solution, this is what happens... Conversion: 175/348 Seconds |========== | 50% Conversion: 179/348 Seconds |========== | 52% Conversion: 183/348 Seconds |========== | 54% Conversion: 187/348 Seconds |=========== | 56% Code I use is... print "Conversion: $converted_seconds/$total_time Seconds $progress_bar $converted_percentage%\n"; I am doing this in Linux using PHP(only I will use the app - so please excuse the language choice). So, the solution should work on the Linux platform - but if you have a solution that's cross platform, that would be preferable.

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• How to list all users in a Linux group?

  - by helix84

  How do I list all members of a group in Linux (and possibly other unices)?

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• Removing old kernel entries in Grub

  - by To Do

  I regularly delete old kernels leaving only the latest two entries using Synaptic. I'm using Precise. However in my Grub "previous Linux version" menu there are quite a few entries labelled 2.6.8. I cannot find these linux-images in Synaptic. dpkg -l | grep linux-image Gives: rc linux-image-3.0.0-17-generic 3.0.0-17.30 Linux kernel image for version 3.0.0 on x86/x86_64 ii linux-image-3.2.0-27-generic 3.2.0-27.43 Linux kernel image for version 3.2.0 on 32 bit x86 SMP ii linux-image-3.2.0-29-generic 3.2.0-29.46 Linux kernel image for version 3.2.0 on 32 bit x86 SMP ii linux-image-3.4.0-030400-generic 3.4.0-030400.201205210521 Linux kernel image for version 3.4.0 on 32 bit x86 SMP ii linux-image-generic 3.2.0.29.31 Generic Linux kernel image Sudo update-grub gives: Generating grub.cfg ... Found linux image: /boot/vmlinuz-3.4.0-030400-generic Found initrd image: /boot/initrd.img-3.4.0-030400-generic Found linux image: /boot/vmlinuz-3.2.0-29-generic Found initrd image: /boot/initrd.img-3.2.0-29-generic Found linux image: /boot/vmlinuz-3.2.0-27-generic Found initrd image: /boot/initrd.img-3.2.0-27-generic Found linux image: /boot/vmlinuz-2.6.38-11-generic Found initrd image: /boot/initrd.img-2.6.38-11-generic Found linux image: /boot/vmlinuz-2.6.38-10-generic Found initrd image: /boot/initrd.img-2.6.38-10-generic Found linux image: /boot/vmlinuz-2.6.38-8-generic Found initrd image: /boot/initrd.img-2.6.38-8-generic Found memtest86+ image: /boot/memtest86+.bin Found Windows Vista (loader) on /dev/sda1 sudo apt-get remove linux-image-2.6.8-8-generic gives: E: Unable to locate package linux-image-2.6.8-8-generic E: Couldn't find any package by regex 'linux-image-2.6.8-8-generic' My boot folder contains the following: abi-2.6.38-10-generic initrd.img-3.4.0-030400-generic abi-2.6.38-11-generic memtest86+.bin abi-2.6.38-8-generic memtest86+_multiboot.bin abi-3.2.0-27-generic System.map-2.6.38-10-generic abi-3.2.0-29-generic System.map-2.6.38-11-generic abi-3.4.0-030400-generic System.map-2.6.38-8-generic config-2.6.38-10-generic System.map-3.2.0-27-generic config-2.6.38-11-generic System.map-3.2.0-29-generic config-2.6.38-8-generic System.map-3.4.0-030400-generic config-3.2.0-27-generic vmcoreinfo-2.6.38-10-generic config-3.2.0-29-generic vmcoreinfo-2.6.38-11-generic config-3.4.0-030400-generic vmcoreinfo-2.6.38-8-generic extlinux vmlinuz-2.6.38-10-generic grub vmlinuz-2.6.38-11-generic initrd.img-2.6.38-10-generic vmlinuz-2.6.38-8-generic initrd.img-2.6.38-11-generic vmlinuz-3.2.0-27-generic initrd.img-2.6.38-8-generic vmlinuz-3.2.0-29-generic initrd.img-3.2.0-27-generic vmlinuz-3.4.0-030400-generic initrd.img-3.2.0-29-generic and ls -l /etc/grub.d yields: total 56 -rwxr-xr-x 1 root root 6715 Apr 17 20:16 00_header -rwxr-xr-x 1 root root 5522 Oct 1 2011 05_debian_theme -rwxr-xr-x 1 root root 7407 May 17 09:22 10_linux -rwxr-xr-x 1 root root 6335 Apr 17 20:16 20_linux_xen -rwxr-xr-x 1 root root 1588 May 3 2011 20_memtest86+ -rwxr-xr-x 1 root root 7603 Apr 17 20:16 30_os-prober -rwxr-xr-x 1 root root 214 Oct 1 2011 40_custom -rwxr-xr-x 1 root root 95 Oct 1 2011 41_custom -rw-r--r-- 1 root root 483 Oct 1 2011 README gdisk -l /dev/sda yields: Partition table scan: MBR: MBR only BSD: not present APM: not present GPT: not present *************************************************************** Found invalid GPT and valid MBR; converting MBR to GPT format. *************************************************************** Disk /dev/sda: 312581808 sectors, 149.1 GiB Logical sector size: 512 bytes Disk identifier (GUID): F832A498-05E1-4615-B5B1-757ACB4A757A Partition table holds up to 128 entries First usable sector is 34, last usable sector is 312581774 Partitions will be aligned on 2048-sector boundaries Total free space is 4183661 sectors (2.0 GiB) Number Start (sector) End (sector) Size Code Name 1 2048 61442047 29.3 GiB 0700 Microsoft basic data 3 163842048 169986047 2.9 GiB 8200 Linux swap 4 169986048 312578047 68.0 GiB 0700 Microsoft basic data 5 61444096 159666175 46.8 GiB 8300 Linux filesystem Please help with removing the old and inexistent kernels from Grub.

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• How To Quickly Reboot Directly from Windows 7 to XP, Vista, or Ubuntu

  - by The Geek

  One of the biggest annoyances with a dual-boot system is having to wait for your PC to reboot to select the operating system you want to switch to, but there’s a simple piece of software that can make this process easier. This guest article was written by Ryan Dozier from the Doztech tech blog. With a small piece of software called iReboot we can skip the above step all together and instantly reboot into the operating system we want right from Windows. Their description says: “Instead of pressing restart, waiting for Windows to shut down, waiting for your BIOS to post, then selecting the operating system you want to boot into (within the bootloader time-limit!); you just select that entry from iReboot and let it do the rest!” Don’t worry about iReboot reconfiguring  your bootloader or any dual boot configuration you have. iReboot will only boot the selected operating system once and go back to your default settings. Using iReboot iReboot is quick and easy to install. Just download it, link below, run through the setup and select the default configuration. iReboot will automatically figure out what operating systems you have installed and appear in the taskbar. Go over to the taskbar and right click on the iReboot icon and select which operating system you want to reboot into. This method will add a check mark on the operating system you want to boot into. On your next reboot the system will automatically load your choice and skip the Windows Boot Manager. If you want to reboot automatically just select “Reboot on Selection” in the iReboot menu.   To be even more productive, you can install iReboot into each Windows operating system to quickly access the others with a few simple clicks.   iReboot does not work in Linux so you will have to reboot manually. Then wait for the Windows Boot Manager to load and select your operating system.   Conclusion iReboot works on  Windows XP, Windows Vista,  and Windows 7 as well as 64 bit versions of these operating systems. Unfortunately iReboot is only available for Windows but you can still use its functionality in Windows to quickly boot up your Linux machine. A simple reboot in Linux will take you back to Windows Boot Manager. Download iReboot from neosmart.net Editor’s note: We’ve not personally tested this software over at How-To Geek, but Neosmart, the author of the software, generally makes quality stuff. Still, you might want to test it out on a test machine first. If you’ve got any experience with this software, please be sure to let your fellow readers know in the comments. Similar Articles Productive Geek Tips Restart the Ubuntu Gnome User Interface QuicklyKeyboard Ninja: 21 Keyboard Shortcut ArticlesTest Your Computer’s Memory Using Windows Vista Memory Diagnostic ToolEnable or Disable UAC From the Windows 7 / Vista Command LineSet Windows as Default OS when Dual Booting Ubuntu 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 Home Networks – How do they look like & the problems they cause Check Your IMAP Mail Offline In Thunderbird Follow Finder Finds You Twitter Users To Follow Combine MP3 Files Easily QuicklyCode Provides Cheatsheets & Other Programming Stuff Download Free MP3s from Amazon

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• multi-user rvm gem install failure when called from CloudFormation::Init

  - by Peter Mounce

  I've taken an Amazon Linux AMI (based on CentOS) and installed RVM (1.10.3) to it in multi-user fashion (see {1} below). I used that to install ruby 1.9.3-p125, rubygems 1.8.17, and bundler 1.1 as the baseline requirements for most things I'm going to be using the instances for. I've captured that instance to an AMI, and am now launching it via CloudFormation, with some CloudFormation::Init commands. One of them is to use s3cmd to pull down a private gem from S3, and the next one, the one that fails, is to install that gem. It fails with an error message 2012-03-15 16:53:20,201 [ERROR] Command 20_install_gems (/usr/local/rvm/rubies/ruby-1.9.3-p125/bin/gem install ./*.gem) failed 2012-03-15 16:53:20,202 [DEBUG] Command 20_install_gems output: /usr/local/rvm/rubies/ruby-1.9.3-p125/bin/gem:12:in `require': no such file to load -- rubygems (LoadError) from /usr/local/rvm/rubies/ruby-1.9.3-p125/bin/gem:12 Now, that happens during the cfn-init execution - I assume, but haven't checked yet, that cfn-init is being run with an environment different from that of ec2-user (there are no other users on the instance). If I run gem install mygem.gem in an interactive session then that works fine. So, my question really, is what should I do to make this work for cfn-init? Have I correctly set up rvm as multi-user? I've confirmed that cfn-init is being run as the root user, with his restricted environment. How should I source the /etc/profile.d/rvm.sh into root's sessions? {1} My semi-automated rvm installation steps (run in interactive session as ec2-user): sudo bash -s stable < <(curl -s https://raw.github.com/wayneeseguin/rvm/master/binscripts/rvm-installer ) sudo gpasswd -a ec2-user rvm # iconv-devel is baked into centos' glibc sudo yum install -y autoconf automake bison bzip2 gcc-c++ git libffi-devel libtool libxml2-devel libxslt-devel libyaml-devel make openssl-devel patch readline readline-devel zlib zlib-devel source /etc/profile.d/rvm.sh rvm list known # in a new session: rvm install ruby-1.9.3-p125 rvm use 1.9.3 --default gem update --system # gems required by public_web-awareness gem install aws-sdk bundler cocaine sinatra echo -e "gem: --no-ri --no-rdoc\n" > /home/ec2-user/.gemrc # delete unnecessary documentation files rm -rf `gem env gemdir`/doc sudo -s sudo echo -e "gem: --no-ri --no-rdoc\n" > /etc/skel/.gemrc sudo echo -e "gem: --no-ri --no-rdoc\n" > /etc/gemrc # ctrl + d out of the sudo session Some environment information: [ec2-user@ip ~]$ echo $PATH /usr/local/rvm/gems/ruby-1.9.3-p125/bin:/usr/local/rvm/gems/ruby-1.9.3-p125@global/bin:/usr/local/rvm/rubies/ruby-1.9.3-p125/bin:/usr/local/rvm/bin:/usr/local/bin:/bin:/usr/bin:/usr/local/sbin:/usr/sbin:/sbin:/opt/aws/bin:/home/ec2-user/bin [ec2-user@ip ~]$ echo $GEM_HOME /usr/local/rvm/gems/ruby-1.9.3-p125 [ec2-user@ip ~]$ echo $GEM_PATH /usr/local/rvm/gems/ruby-1.9.3-p125:/usr/local/rvm/gems/ruby-1.9.3-p125@global [ec2-user@ip ~]$ echo $BUNDLE_PATH [ec2-user@ip ~]$ gem list *** LOCAL GEMS *** aws-sdk (1.3.6) bundler (1.1.0) cocaine (0.2.1) httparty (0.8.1) json (1.6.5) multi_json (1.1.0) multi_xml (0.4.1) nokogiri (1.5.1, 1.5.0) rack (1.4.1) rack-protection (1.2.0) rake (0.9.2) sinatra (1.3.2) tilt (1.3.3) uuidtools (2.1.2) yamler (0.1.0)

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• Django User M2M relationship

  - by Antonio

  When trying to syncdb with the following models: class Contact(models.Model): user_from = models.ForeignKey(User,related_name='from_user') user_to = models.ForeignKey(User, related_name='to_user') class Meta: unique_together = (('user_from', 'user_to'),) User.add_to_class('following', models.ManyToManyField('self', through=Contact, related_name='followers', symmetrical=False)) I get the following error: Error: One or more models did not validate: auth.user: Accessor for m2m field 'following' clashes with related m2m field 'User.followers'. Add a related_name argument to the definition for 'following'. auth.user: Reverse query name for m2m field 'following' clashes with related m2m field 'User.followers'. Add a related_name argument to the definition for 'following'. auth.user: The model User has two manually-defined m2m relations through the model Contact, which is not permitted. Please consider using an extra field on your intermediary model instead. auth.user: Accessor for m2m field 'following' clashes with related m2m field 'User.followers'. Add a related_name argument to the definition for 'following'. auth.user: Reverse query name for m2m field 'following' clashes with related m2m field 'User.followers'. Add a related_name argument to the definition for 'following'.

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• Have an unprivileged non-account user ssh into another box?

  - by Daniel Quinn

  I know how to get a user to ssh into another box with a key: ssh -l targetuser -i path/to/key targethost But what about non-account users like apache? As this user doesn't have a home directory to which it can write a .ssh directory, the whole thing keeps failing with: $ sudo -u apache ssh -o StrictHostKeyChecking=no -l targetuser -i path/to/key targethost Could not create directory '/var/www/.ssh'. Warning: Permanently added '<hostname>' (RSA) to the list of known hosts. Permission denied (publickey). I've tried variations using -o UserKnownHostsFile=/dev/null and setting $HOME to /dev/null and none of these have done the trick. I understand that sudo could probably fix this for me, but I'm trying to avoid having to require a manual server config since this code will be deployed on a number of different environments. Any ideas? Here's a few examples of what I've tried that don't work: $ sudo -u apache export HOME=path/to/apache/writable/dir/ ssh -o StrictHostKeyChecking=no -o UserKnownHostsFile=path/to/apache/writable/dir/.ssh/known_hosts -l deploy -i path/to/key targethost $ sudo -u apache ssh -o StrictHostKeyChecking=no -o UserKnownHostsFile=path/to/apache/writable/dir/.ssh/known_hosts -l deploy -i path/to/key targethost $ sudo -u apache ssh -o StrictHostKeyChecking=no -o UserKnownHostsFile=/dev/null -l deploy -i path/to/key targethost Eventually, I'll be using this solution to run rsync as the apache user.

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• Windows 7 / Ubuntu Dualboot GRUB Problem.

  - by Tek

  I'd like to first say ahead of time that I'm running a RAID-0 Setup. 1.First of all, I'm glad Ubuntu 9.10 installed flawlessly and detected my RAID-0 setup just fine. The issue I'm having now is that I already had Windows 7 installed and made a small 12GB partition for Linux/Swap. I grabbed EasyBCD 2.0 to edit the W7 bootloader and configured it to use dual boot Grub2 because before it didn't even show the option for Ubuntu. The bootloader points to a file made in the windows directory made by EasyBCD called "C:\NST\AutoNeoGrub0.mbr" which is what I'm guessing grub is booting from. After that I got the option for booting Ubuntu. The problem is that it's sending me to the Grub prompt (probably because it's pointing to \NST|AutoNeoGrub0.mbr?), at first I didn't know what to do but I researched and have to type grub commands to manually boot into Ubuntu Linux. Ex: grubroot (hd0,4) grubkernel /boot/vmlinuz-2.6... root=/dev/disk/by-uuid/24624-2424... grubinitrd boot/initrd.img-2.6... grubboot After all that Ubuntu boots just fine, but how do I fix it permanently? Do I need to edit the bootloader manually (since Easy BCD "autoconfigures")? Some insight on this would rock! Also, it sucks to type the actual uuid since it's REALLY long. I tried getting the name of the drive via fdisk -l but since it's raid 0 I'm guessing I can't do that. How can I get a shorter name of the drive? like /dev/sda, /dev/sdb etc? I've also tried to update to the latest GRUB and I got this: Creating config file /etc/default/grub with new version Generating core.img error: cannot seek /dev/sdc' error: cannot seek/dev/sdc' grub-probe: error: no mapping exists for nvidia_dbedfcca5' Auto-detection of a filesystem module failed. Please specify the module with the option--modules' explicitly. dpkg: error processing grub-pc (--configure): subprocess installed post-installation script returned error exit status 1 dpkg: dependency problems prevent configuration of grub2: grub2 depends on grub-pc; however: Package grub-pc is not configured yet. dpkg: error processing grub2 (--configure): dependency problems - leaving unconfigured No apport report written because the error message indicates its a followup error from a previous failure. E: Sub-process /usr/bin/dpkg returned an error code (1) I've also tried: b@dnb:~$ sudo update-grub error: cannot seek /dev/sdc' error: cannot seek/dev/sdc' Generating grub.cfg ... Found linux image: /boot/vmlinuz-2.6.31-14-generic Found initrd image: /boot/initrd.img-2.6.31-14-generic error: cannot seek /dev/sdc' grub-probe: error: no mapping exists fornvidia_dbedfcca5' error: cannot seek /dev/sdc' grub-probe: error: no mapping exists fornvidia_dbedfcca5' Found memtest86+ image: /boot/memtest86+.bin Found Windows 7 (loader) on /dev/mapper/nvidia_dbedfcca1 error: cannot seek /dev/sdc' grub-probe: error: no mapping exists fornvidia_dbedfcca1' done To no avail. Any idea what I can do to fix this mess? :( Edit: This is my disk configuration. b@dnb:~$ sudo df -l Filesystem 1K-blocks Used Available Use% Mounted on /dev/mapper/nvidia_dbedfcca5 12302232 2744788 8932520 24% / udev 1030288 268 1030020 1% /dev none 1030288 964 1029324 1% /dev/shm none 1030288 92 1030196 1% /var/run none 1030288 0 1030288 0% /var/lock none 1030288 0 1030288 0% /lib/init/rw /dev/sr0 706532 706532 0 100% /media/cdrom0 Note: /dev/mapper/nvidia_dbedfcca5 is my Linux boot partition

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• How to get Alfresco login ticket without user password, but with impersonating user with user principal name (UPN)

  - by dok

  I'm writing a DLL that has function for getting Alfresco login ticket without using user password, using only a user principal name (UPN). I’m calling alfresco REST API service /wcservice. I use NTLM in Alfresco. I’m impersonating users using WindowsIdentity constructor as explained here http://msdn.microsoft.com/en-us/library/ms998351.aspx#paght000023_impersonatingbyusingwindowsidentity. I checked and user is properly impersonated (I checked WindowsIdentity.GetCurrent().Name property). After impersonating a user, I try to make HttpWebRequest and set its credentials with CredentialsCache.DefaultNetworkCredentials. I get the error: The remote server returned an error: (401) Unauthorized. at System.Net.HttpWebRequest.GetResponse() When I use new NetworkCredential("username", "P@ssw0rd") to set request credentials, I get Alfresco login ticket (HttpStatusCode.OK, 200). Is there any way that I can get Alfresco login ticket without user password? Here is the code that I'm using: private string GetTicket(string UPN) { WindowsIdentity identity = new WindowsIdentity(UPN); WindowsImpersonationContext context = null; try { context = identity.Impersonate(); MakeWebRequest(); } catch (Exception e) { return e.Message + Environment.NewLine + e.StackTrace; } finally { if (context != null) { context.Undo(); } } } private string MakeWebRequest() { string URI = "http://alfrescoserver/alfresco/wcservice/mg/util/login"; HttpWebRequest request = WebRequest.Create(URI) as HttpWebRequest; request.CookieContainer = new CookieContainer(1); //request.Credentials = new NetworkCredential("username", "p@ssw0rd"); // It works with this request.Credentials = CredentialCache.DefaultNetworkCredentials; // It doesn’t work with this //request.Credentials = CredentialCache.DefaultCredentials; // It doesn’t work with this either try { using (HttpWebResponse response = request.GetResponse() as HttpWebResponse) { StreamReader sr = new StreamReader(response.GetResponseStream()); return sr.ReadToEnd(); } } catch (Exception e) { return (e.Message + Environment.NewLine + e.StackTrace); } } Here are records from Alfresco stdout.log (if it helps in any way): 17:18:04,550 DEBUG [app.servlet.NTLMAuthenticationFilter] Processing request: /alfresco/wcservice/mg/util/login SID:7453F7BD4FD2E6A61AD40A31A37733A5 17:18:04,550 DEBUG [web.scripts.DeclarativeRegistry] Web Script index lookup for uri /mg/util/login took 0.526239ms 17:18:04,550 DEBUG [app.servlet.NTLMAuthenticationFilter] New NTLM auth request from 10.**.**.** (10.**.**.**:1229) 17:18:04,566 DEBUG [app.servlet.NTLMAuthenticationFilter] Processing request: /alfresco/wcservice/mg/util/login SID:7453F7BD4FD2E6A61AD40A31A37733A5 17:18:04,566 DEBUG [web.scripts.DeclarativeRegistry] Web Script index lookup for uri /mg/util/login took 0.400909ms 17:18:04,566 DEBUG [app.servlet.NTLMAuthenticationFilter] Received type1 [Type1:0xe20882b7,Domain:<NotSet>,Wks:<NotSet>] 17:18:04,566 DEBUG [app.servlet.NTLMAuthenticationFilter] Client domain null 17:18:04,675 DEBUG [app.servlet.NTLMAuthenticationFilter] Sending NTLM type2 to client - [Type2:0x80000283,Target:AlfrescoServerA,Ch:197e2631cc3f9e0a]

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