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  • Combine config-paramters with parameters passed from commanline

    - by Frederik
    I have created a SSIS-package that imports a file into a table (simple enough). I have some variables, a few set in a config-file such as server, database, importfolder. at runtime I want to pass the filename. This is done through a stored procedure using dtexec. When setting the paramters throught the configfile it works fine also when setting all parameters in the procedure and passing them with the \Set statement (se below). when I try to combine the config-version with settings parameters on the fly I get an error refering to the config-files path that was set at design time. Has anybody come across this and found a solution for it? Regards Frederik DECLARE @SSISSTR VARCHAR(8000), @DataBaseServer VARCHAR(100), @DataBaseName VARCHAR(100), @PackageFilePath VARCHAR(200), @ImportFolder VARCHAR(200), @HandledFolder VARCHAR(200), @ConfigFilePath VARCHAR(200), @SSISreturncode INT; /* DEBUGGING DECLARE @FileName VARCHAR(100), @SelectionId INT SET @FileName = 'Test.csv'; SET @SelectionId = 366; */ SET @PackageFilePath = '/FILE "Y:\SSIS\Packages\PostalCodeSelectionImport\ImportPackage.dtsx" '; SET @DataBaseServer = 'STOSWVUTVDB01\DEV_BSE'; SET @DataBaseName = 'BSE_ODR'; SET @ImportFolder = '\\Stoswvutvbse01\Application\FileLoadArea\ODR\\'; SET @HandledFolder = '\\Stoswvutvbse01\Application\FileLoadArea\ODR\Handled\\'; --SET @ConfigFilePath = '/CONFIGFILE "Y:\SSIS\Packages\PostalCodeSelectionImport\Configuration\DEV_BSE.dtsConfig" '; ----now making "dtexec" SQL from dynamic values SET @SSISSTR = 'DTEXEC ' + @PackageFilePath; -- + @ConfigFilePath; SET @SSISSTR = @SSISSTR + ' /SET \Package.Variables[User::SelectionId].Properties[Value];' + CAST( @SelectionId AS VARCHAR(12)); SET @SSISSTR = @SSISSTR + ' /SET \Package.Variables[User::DataBaseServer].Properties[Value];"' + @DataBaseServer + '"'; SET @SSISSTR = @SSISSTR + ' /SET \Package.Variables[User::ImportFolder].Properties[Value];"' + @ImportFolder + '" '; SET @SSISSTR = @SSISSTR + ' /SET \Package.Variables[User::DataBaseName].Properties[Value];"' + @DataBaseName + '" '; SET @SSISSTR = @SSISSTR + ' /SET \Package.Variables[User::ImportFileName].Properties[Value];"' + @FileName + '" '; SET @SSISSTR = @SSISSTR + ' /SET \Package.Variables[User::HandledFolder].Properties[Value];"' + @HandledFolder + '" '; -- Now execute dynamic SQL by using EXEC. EXEC @SSISreturncode = xp_cmdshell @SSISSTR;

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  • Java Builder pattern with Generic type bounds

    - by I82Much
    Hi all, I'm attempting to create a class with many parameters, using a Builder pattern rather than telescoping constructors. I'm doing this in the way described by Joshua Bloch's Effective Java, having private constructor on the enclosing class, and a public static Builder class. The Builder class ensures the object is in a consistent state before calling build(), at which point it delegates the construction of the enclosing object to the private constructor. Thus public class Foo { // Many variables private Foo(Builder b) { // Use all of b's variables to initialize self } public static final class Builder { public Builder(/* required variables */) { } public Builder var1(Var var) { // set it return this; } public Foo build() { return new Foo(this); } } } I then want to add type bounds to some of the variables, and thus need to parametrize the class definition. I want the bounds of the Foo class to be the same as that of the Builder class. public class Foo<Q extends Quantity> { private final Unit<Q> units; // Many variables private Foo(Builder<Q> b) { // Use all of b's variables to initialize self } public static final class Builder<Q extends Quantity> { private Unit<Q> units; public Builder(/* required variables */) { } public Builder units(Unit<Q> units) { this.units = units; return this; } public Foo build() { return new Foo<Q>(this); } } } This compiles fine, but the compiler is allowing me to do things I feel should be compiler errors. E.g. public static final Foo.Builder<Acceleration> x_Body_AccelField = new Foo.Builder<Acceleration>() .units(SI.METER) .build(); Here the units argument is not Unit<Acceleration> but Unit<Length>, but it is still accepted by the compiler. What am I doing wrong here? I want to ensure at compile time that the unit types match up correctly.

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  • Accessing py2exe program over network in Windows 98 throws ImportErrors

    - by darvids0n
    I'm running a py2exe-compiled python program from one server machine on a number of client machines (mapped to a network drive on every machine, say W:). For Windows XP and later machines, have so far had zero problems with Python picking up W:\python23.dll (yes, I'm using Python 2.3.5 for W98 compatibility and all that). It will then use W:\zlib.pyd to decompress W:\library.zip containing all the .pyc files like os and such, which are then imported and the program runs no problems. The issue I'm getting is on some Windows 98 SE machines (note: SOME Windows 98 SE machines, others seem to work with no apparent issues). What happens is, the program runs from W:, the W:\python23.dll is, I assume, found (since I'm getting Python ImportErrors, we'd need to be able to execute a Python import statement), but a couple of things don't work: 1) If W:\library.zip contains the only copy of the .pyc files, I get ZipImportError: can't decompress data; zlib not available (nonsense, considering W:\zlib.pyd IS available and works fine with the XP and higher machines on the same network). 2) If the .pyc files are actually bundled INSIDE the python exe by py2exe, OR put in the same directory as the .exe, OR put into a named subdirectory which is then set as part of the PYTHONPATH variable (e.g W:\pylib), I get ImportError: no module named os (os is the first module imported, before sys and anything else). Come to think of it, sys.path wouldn't be available to search if os was imported before it maybe? I'll try switching the order of those imports but my question still stands: Why is this a sporadic issue, working on some networks but not on others? And how would I force Python to find the files that are bundled inside the very executable I run? I have immediate access to the working Windows 98 SE machine, but I only get access to the non-working one (a customer of mine) every morning before their store opens. Thanks in advance! EDIT: Okay, big step forward. After debugging with PY2EXE_VERBOSE, the problem occurring on the specific W98SE machine is that it's not using the right path syntax when looking for imports. Firstly, it doesn't seem to read the PYTHONPATH environment variable (there may be a py2exe-specific one I'm not aware of, like PY2EXE_VERBOSE). Secondly, it only looks in one place before giving up (if the files are bundled inside the EXE, it looks there. If not, it looks in library.zip). EDIT 2: In fact, according to this, there is a difference between the sys.path in the Python interpreter and that of Py2exe executables. Specifically, sys.path contains only a single entry: the full pathname of the shared code archive. Blah. No fallbacks? Not even the current working directory? I'd try adding W:\ to PATH, but py2exe doesn't conform to any sort of standards for locating system libraries, so it won't work. Now for the interesting bit. The path it tries to load atexit, os, etc. from is: W:\\library.zip\<module>.<ext> Note the single slash after library.zip, but the double slash after the drive letter (someone correct me if this is intended and should work). It looks like if this is a string literal, then since the slash isn't doubled, it's read as an (invalid) escape sequence and the raw character is printed (giving W:\library.zipos.pyd, W:\library.zipos.dll, ... instead of with a slash); if it is NOT a string literal, the double slash might not be normpath'd automatically (as it should be) and so the double slash confuses the module loader. Like I said, I can't just set PYTHONPATH=W:\\library.zip\\ because it ignores that variable. It may be worth using sys.path.append at the start of my program but hard-coding module paths is an absolute LAST resort, especially since the problem occurs in ONE configuration of an outdated OS. Any ideas? I have one, which is to normpath the sys.path.. pity I need os for that. Another is to just append os.getenv('PATH') or os.getenv('PYTHONPATH') to sys.path... again, needing the os module. The site module also fails to initialise, so I can't use a .pth file. I also recently tried the following code at the start of the program: for pth in sys.path: fErr.write(pth) fErr.write(' to ') pth.replace('\\\\','\\') # Fix Windows 98 pathing issues fErr.write(pth) fErr.write('\n') But it can't load linecache.pyc, or anything else for that matter; it can't actually execute those commands from the looks of things. Is there any way to use built-in functionality which doesn't need linecache to modify the sys.path dynamically? Or am I reduced to hard-coding the correct sys.path?

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  • Objects in Java ArrayList don't get updated.

    - by Sbm007
    This is going to be a very long post, hopefully you can understand what I'm talking about and I appreciate any help. Thanks Basically, I've created a personal, non-commercial project (which I don't plan to release) that can read ZIP and RAR files. It can only read the contents in the archive, the folders inside, the files inside the folders and its properties (such as last modified date, last modified time, CRC checksum, uncompressed size, compressed size and file name). It can't extract files either, so it's really a ZIP/RAR viewer if you may. Anyway that's slightly irrelevant to my problem but I thought I'd give you some background info. Now for my problem: I can successfully list all the folders and files inside a ZIP archive, so now I want to take that raw input and link it together in some useful way. I made 2 classes: ArchiveFile (represents a file inside a ZIP) and ArchiveFolder (represents a folder inside a ZIP). They both have some useful methods such as getLastModifiedDate, getName, getPath and so on. But the difference is that ArchiveFolder can hold an ArrayList of ArchiveFile's and additional ArchiveFolder's (think of this as files and folders inside a folder). Now I want to populate my raw input into one root ArchiveFolder, which will have all the files in the root dir of the ZIP in the ArchiveFile's ArrayList and any additional folders in the root dir of the ZIP in the ArchiveFolder's ArrayList (and this process can continue on like this like a chain reaction (more files/folders in that ArchiveFolder etc etc). So I came up with the following code: while (archive.hasMore()) { String path = ""; ArchiveFolder current = root; String[] contents = archive.getName().split("/"); for (int x = 0; x < contents.length; ++x) { if (x == (contents.length - 1) && !archive.getName().endsWith("/")) { // If on last item and item is a file path += contents[x]; // Update final path ArchiveFile file = new ArchiveFile(path, contents[x], archive.getUncompressedSize(), archive.getCompressedSize(), archive.getModifiedTime(), archive.getModifiedDate(), archive.getCRC()); current.addFile(file); // Create and add the file to the current ArchiveFolder } else if (x == (contents.length - 1)) { // Else if we are on last item and it is a folder path += contents[x] + "/"; // Update final path ArchiveFolder folder = new ArchiveFolder(path, contents[x], archive.getModifiedTime(), archive.getModifiedDate()); current.addFolder(folder); // Create and add this folder to the current ArchiveFile } else { // Else if we are still traversing through the path path += contents[x] + "/"; // Update path ArchiveFolder folder = new ArchiveFolder(path, contents[x]); current.addFolder(folder); // Create and add folder (remember we do not know the modified date/time as all we know is the path, so we can deduce the name only) current = folder; // Update current ArchiveFolder to the newly created one for the next iteration of the for loop } } archive.getNext(); } Assume that root is the root ArchiveFolder (initially empty). And that archive.getName() returns the name of the current file OR folder in the following fashion: file.txt or folder1/file2.txt or folder4/folder2/ (this is a empty folder) etc. So basically the relative path from the root of the ZIP archive. Please read through the comments in the above code to familiarize yourself with it. Also assume that the addFolder method in an ArchiveFile, only adds the folder if it doesn't exist already (so there are no multiple folders) and it also updates the time and date of an existing folder if it is blank (ie it was a intermediate folder we only knew the name of, but now we know its details). The code for addFolder is (pretty self-explanitory): public void addFolder(ArchiveFolder folder) { int loc = folders.indexOf(folder); // folders is the ArrayList containing ArchiveFolder's if (loc == -1) { folders.add(folder); } else { ArchiveFolder real = folders.get(loc); if (real.time == null) { real.setTime(folder.getTime()); real.setDate(folder.getDate()); } } } So I can't see anything wrong with the code, it works and after finishing, the root ArchiveFolder contains all the files in the root of the ZIP as I want it to, and it contains all the direcories in the root folder as I want it to. So you'd think it works as expected, but no the ArchiveFolder's in the root folder don't contain the data inside those 'child' folders, it's just a blank folder with no additional files and folders (while it does really contain some more files/folders when viewed in WinZip). After debugging using Eclipse, the for loop does iterate through all the files (even those not included above), so this led me to believe that there is a problem with this line of the code: current = folder; What it does is, it updates the current folder (used as an intermediate by the loop) to the newly added folder. I thought Java passed by reference and thus all new operations and new additions in future ArchiveFile's and ArchiveFolder's are automatically updated, and parent ArchiveFolder's will be updated accordingly. But that does not appear to be the case? I know this is a long ass post and I really hope anyone can help me out with this. Thanks in advance.

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  • How to Animate Text and Objects in PowerPoint 2010

    - by DigitalGeekery
    Are you looking for an eye catching way to keep your audience interested in your PowerPoint presentations? Today we’ll take a look at how to add animation effects to objects in PowerPoint 2010. Select the object you wish to animate and then click the More button in the Animation group of the Animation tab.   Animations are grouped into four categories. Entrance effects, Exit effects, Emphasis effects, and Motion Paths. You can get a Live Preview of how the animation will look by hovering your mouse over an animation effect.   When you select a Motion Path, your object will move along the dashed path line as shown on the screen. (This path is not displayed in the final output) Certain aspects of the Motion Path effects are editable. When you apply a Motion Path animation to an object, you can select the path and drag the end to change the length or size of the path. The green marker along the motion path marks the beginning of the  path and the red marks the end. The effects can be rotated by clicking and the bar near the center of the effect.   You can display additional effects by choosing one of the options at the bottom. This will pop up a Change Effect window. If you have Preview Effect checked at the lower left you can preview the effects by single clicking.   Apply Multiple Animations to an Object Select the object and then click the Add Animation button to display the animation effects. Just as we did with the first effect, you can hover over to get a live preview. Click to apply the effect. The animation effects will happen in the order they are applied. Animation Pane You can view a list of the animations applied to a slide by opening the Animation Pane. Select the Animation Pane button from the Advanced Animation group to display the Animation Pane on the right. You’ll see that each animation effect in the animation pane has an assigned number to the left.    Timing Animation Effects You can change when your animation starts to play. By default it is On Click. To change it, select the effect in the Animation Pane and then choose one of the options from the Start dropdown list. With Previous starts at the same time as the previous animation and After Previous starts after the last animation. You can also edit the duration that the animations plays and also set a delay.   You can change the order in which the animation effects are applied by selecting the effect in the animation pane and clicking Move Earlier or Move Later from the Timing group on the Animation tab. Effect Options If the Effect Options button is available when your animation is selected, then that particular animation has some additional effect settings that can be configured. You can access the Effect Option by right-clicking on the the animation in the Animation Pane, or by selecting Effect Options on the ribbon.   The available options will vary by effect and not all animation effects will have Effect Options settings. In the example below, you can change the amount of spinning and whether the object will spin clockwise or counterclockwise.   Under Enhancements, you can add sound effects to your animation. When you’re finished click OK.   Animating Text Animating Text works the same as animating an object. Simply select your text box and choose an animation. Text does have some different Effect Options. By selecting a sequence, you decide whether the text appears as one object, all at once, or by paragraph. As is the case with objects, there will be different available Effect Options depending on the animation you choose. Some animations, such as the Fly In animation, will have directional options.   Testing Your Animations Click on the Preview button at any time to test how your animations look. You can also select the Play button on the Animation Pane. Conclusion Animation effects are a great way to focus audience attention on important points and hold viewers interest in your PowerPoint presentations. Another cool way to spice up your PPT 2010 presentations is to add video from the web. What tips do you guys have for making your PowerPoint presentations more interesting? Similar Articles Productive Geek Tips Center Pictures and Other Objects in Office 2007 & 2010Preview Before You Paste with Live Preview in Office 2010Embed True Type Fonts in Word and PowerPoint 2007 DocumentsHow to Add Video from the Web in PowerPoint 2010Add Artistic Effects to Your Pictures in Office 2010 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 Xobni Plus for Outlook All My Movies 5.9 CloudBerry Online Backup 1.5 for Windows Home Server Snagit 10 24 Million Sites Windows Media Player Glass Icons (icons we like) How to Forecast Weather, without Gadgets Outlook Tools, one stop tweaking for any Outlook version Zoofs, find the most popular tweeted YouTube videos Video preview of new Windows Live Essentials

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  • Tip #13 java.io.File Surprises

    - by ByronNevins
    There is an assumption that I've seen in code many times that is totally wrong.  And this assumption can easily bite you.  The assumption is: File.getAbsolutePath and getAbsoluteFile return paths that are not relative.  Not true!  Sort of.  At least not in the way many people would assume.  All they do is make sure that the beginning of the path is absolute.  The rest of the path can be loaded with relative path elements.  What do you think the following code will print? public class Main {    public static void main(String[] args) {        try {            File f = new File("/temp/../temp/../temp/../");            File abs  = f.getAbsoluteFile();            File parent = abs.getParentFile();            System.out.println("Exists: " + f.exists());            System.out.println("Absolute Path: " + abs);            System.out.println("FileName: " + abs.getName());            System.out.printf("The Parent Directory of %s is %s\n", abs, parent);            System.out.printf("The CANONICAL Parent Directory of CANONICAL %s is %s\n",                        abs, abs.getCanonicalFile().getParent());            System.out.printf("The CANONICAL Parent Directory of ABSOLUTE %s is %s\n",                        abs, parent.getCanonicalFile());            System.out.println("Canonical Path: " + f.getCanonicalPath());        }        catch (IOException ex) {            System.out.println("Got an exception: " + ex);        }    }} Output: Exists: trueAbsolute Path: D:\temp\..\temp\..\temp\..FileName: ..The Parent Directory of D:\temp\..\temp\..\temp\.. is D:\temp\..\temp\..\tempThe CANONICAL Parent Directory of CANONICAL D:\temp\..\temp\..\temp\.. is nullThe CANONICAL Parent Directory of ABSOLUTE D:\temp\..\temp\..\temp\.. is D:\tempCanonical Path: D:\ Notice how it says that the parent of d:\ is d:\temp !!!The file, f, is really the root directory.  The parent is supposed to be null. I learned about this the hard way! getParentXXX simply hacks off the final item in the path. You can get totally unexpected results like the above. Easily. I filed a bug on this behavior a few years ago[1].   Recommendations: (1) Use getCanonical instead of getAbsolute.  There is a 1:1 mapping of files and canonical filenames.  I.e each file has one and only one canonical filename and it will definitely not have relative path elements in it.  There are an infinite number of absolute paths for each file. (2) To get the parent file for File f do the following instead of getParentFile: File parent = new File(f, ".."); [1] http://bt2ws.central.sun.com/CrPrint?id=6687287

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  • Impersonation in ASP.NET MVC

    - by eibrahim
    I have an Action that needs to read a file from a secure location, so I have to use impersonation to read the file. This code WORKS: [AcceptVerbs(HttpVerbs.Get)] public ActionResult DirectDownload(Guid id) { if (Impersonator.ImpersonateValidUser()) { try { var path = "path to file"; if (!System.IO.File.Exists(path)) { return View("filenotfound"); } var bytes = System.IO.File.ReadAllBytes(path); return File(bytes, "application/octet-stream", "FileName"); } catch (Exception e) { Log.Exception(e); }finally { Impersonator.UndoImpersonation(); } } return View("filenotfound"); } The only problem with the above code is that I have to read the entire file into memory and I am going to be dealing with VERY large files, so this is not a good solution. But if I replace these 2 lines: var bytes = System.IO.File.ReadAllBytes(path); return File(bytes, "application/octet-stream", "FileName"); with this: return File(path, "application/octet-stream", "FileName"); It does NOT work and I get the error message: Access to the path 'c:\projects\uploads\1\aa2bcbe7-ea99-499d-add8-c1fdac561b0e\Untitled 2.csv' is denied. I guess using the File results with a path, tries to open the file at a later time in the request pipeline when I have already "undone" the impersonation. Remember, the impersonation code works because I can read the file in the bytes array. What I want to do though is stream the file to the client. Any idea how I can work around this? Thanks in advance.

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  • Re: How can Django/WSGI and PHP share / on Apache?

    - by Bogdan
    in response to: How can Django/WSGI and PHP share / on Apache? Hello, could you please post the complete config file from /sites-available I am having a problem seems like rewrite engine redirects all requests to django, so static and php files are not served and instead i see the django 404 page. If I get rid of rewrite rule then static files and php works. here is my apache config file from /sites-available <VirtualHost *:80> ServerAdmin webmaster@localhost DocumentRoot /home/www/django <Directory /> Options +FollowSymLinks ExecCGI Indexes AllowOverride None DirectoryIndex index.php AddHandler wsgi-script .wsgi </Directory> RewriteEngine On RewriteCond %{REQUEST_FILENAME} !-f RewriteRule ^(.*)$ /mysite.wsgi/$1 [QSA,PT,L] ~ and my .wsgi file: import site site.addsitedir('/home/user/.virtualenvs/url.com/lib/python2.6/site-packages') import os, sys path = '/home/www/django' if path not in sys.path: sys.path.append(path) os.environ['DJANGO_SETTINGS_MODULE'] = 'mysite.settings' sys.path.append(path + '/mysite') import django.core.handlers.wsgi _application = django.core.handlers.wsgi.WSGIHandler() import posixpath def application(environ, start_response): # Wrapper to set SCRIPT_NAME to actual mount point. environ['SCRIPT_NAME'] = posixpath.dirname(environ['SCRIPT_NAME']) if environ['SCRIPT_NAME'] == '/': environ['SCRIPT_NAME'] = '' return _application(environ, start_response) the document root directory on disk (/home/www/django) contains php files, images, and the mysite.wsgi file.. thanks for your help

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  • Writing files in App_Data causes tempdata to be null

    - by RAMX
    I have a small asp.net MVC 1 web app that can store files and create directories in the App_Data directory. When the write operation succeeds, I add a message to the tempdata and do a redirectToRoute. The problem is that the tempdata is null when the action is executed. If i write the files in a directory outside of the web applications root directory, the tempdata is not null and everything works correctly. Any ideas why writing in the app_data seems to clear the tempdata ? edit: if DRS.Logic.Repository.Manager.CreateFile(path, hpf, comment) writes in the App_Data, TempData will be null in the action being redirected to. if it is a directory out of the web app root it is fine. No exceptions are being thrown. [AcceptVerbs(HttpVerbs.Post)] public ActionResult Create(int id, string path, FormCollection form) { ViewData["path"] = path; ViewData["id"] = id; HttpPostedFileBase hpf; string comment = form["FileComment"]; hpf = Request.Files["File"] as HttpPostedFileBase; if (hpf.ContentLength != 0) { DRS.Logic.Repository.Manager.CreateFile(path, hpf, comment); TempData["notification"] = "file was created"; return RedirectToRoute(new { controller = "File", action ="ViewDetails", id = id, path = path + Path.GetFileName(hpf.FileName) }); } else { TempData["notification"] = "No file were selected."; return View(); } }

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  • Why do I get completely different results when saving a BitmapSource to bmp, jpeg, and png in WPF

    - by DanM
    I wrote a little utility class that saves BitmapSource objects to image files. The image files can be either bmp, jpeg, or png. Here is the code: public class BitmapProcessor { public void SaveAsBmp(BitmapSource bitmapSource, string path) { Save(bitmapSource, path, new BmpBitmapEncoder()); } public void SaveAsJpg(BitmapSource bitmapSource, string path) { Save(bitmapSource, path, new JpegBitmapEncoder()); } public void SaveAsPng(BitmapSource bitmapSource, string path) { Save(bitmapSource, path, new PngBitmapEncoder()); } private void Save(BitmapSource bitmapSource, string path, BitmapEncoder encoder) { using (var stream = new FileStream(path, FileMode.Create)) { encoder.Frames.Add(BitmapFrame.Create(bitmapSource)); encoder.Save(stream); } } } Each of the three Save methods work, but I get unexpected results with bmp and jpeg. Png is the only format that produces an exact reproduction of what I see if I show the BitmapSource on screen using a WPF Image control. Here are the results: BMP - too dark JPEG - too saturated PNG - correct Why am I getting completely different results for different file types? I should note that the BitmapSource in my example uses an alpha value of 0.1 (which is why it appears very desaturated), but it should be possible to show the resulting colors in any image format. I know if I take a screen capture using something like HyperSnap, it will look correct regardless of what file type I save to. Here's a HyperSnap screen capture saved as a bmp: As you can see, this isn't a problem, so there's definitely something strange about WPF's image encoders. Do I have a setting wrong? Am I missing something?

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  • register_globals error in php

    - by user145862
    I was stuck up with the error directive 'register_globals' is no longer available in PHP in unknown on line 0 when tried to check the php version using "php -v" after enabling register_globals in php.ini file. I am not getting any php version info by doing so. Instead it throws the above mentioned error.After turning off this option, php info works quite well. It is very essential for me to have register_globals to be turned on.How can I have this corrected. my php.ini is as follows: ; Default Value: None ; Development Value: "GP" ; Production Value: "GP" ; http://php.net/request-order request_order = "GP" ; Whether or not to register the EGPCS variables as global variables. You may ; want to turn this off if you don't want to clutter your scripts' global scope ; with user data. ; You should do your best to write your scripts so that they do not require ; register_globals to be on; Using form variables as globals can easily lead ; to possible security problems, if the code is not very well thought of. ; register_globals = On ; Determines whether the deprecated long $HTTP_*_VARS type predefined variables ; are registered by PHP or not. As they are deprecated, we obviously don't ; recommend you use them. They are on by default for compatibility reasons but ; they are not recommended on production servers. ; Default Value: On ; Development Value: Off ; Production Value: Off ; register_long_arrays = Off ; This directive determines whether PHP registers $argv & $argc each time it ; runs. $argv contains an array of all the arguments passed to PHP when a script ; is invoked. $argc contains an integer representing the number of arguments ; that were passed when the script was invoked. These arrays are extremely ; useful when running scripts from the command line. When this directive is ; enabled, registering these variables consumes CPU cycles and memory each time ; a script is executed. For performance reasons, this feature should be disabled ; on production servers. ; Note: This directive is hardcoded to On for the CLI SAPI ; Default Value: On ; Development Value: Off ; Production Value: Off ; register_argc_argv = Off ; When enabled, the SERVER and ENV variables are created when they're first ; used (Just In Time) instead of when the script starts. If these variables ; are not used within a script, having this directive on will result in a ; performance gain. The PHP directives register_globals, register_long_arrays, ; and register_argc_argv must be disabled for this directive to have any affect. ; auto_globals_jit = On

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  • Trying to use a authlogic-connect as a plugin in place of gem - Server doesn't start

    - by Arkid
    I am trying to use Authlogic-connect as a plugin in Rails 3 in place of a gem. I have made an entry in the gemfile as gem "authlogic-connect", :require => "authlogic-connect", :path => "localgems" Now when I run the bundle install, it runs fine. When I try to start the server i get the error Could not find gem 'authlogic-connect (>= 0, runtime)' in source at localgems. Source does not contain any versions of 'authlogic-connect (>= 0, runtime)' Try running `bundle install`. I have placed the unzipped Gem renamed as authlogic-connect in the localgems folder. what is the problem? Here is what I get on using rails plugin install arkidmitra$ rails plugin install git://github.com/viatropos/authlogic-connect.git Usage: rails new APP_PATH [options] Options: [--skip-gemfile] # Don't create a Gemfile -d, [--database=DATABASE] # Preconfigure for selected database (options: mysql/oracle/postgresql/sqlite3/frontbase/ibm_db) # Default: sqlite3 -O, [--skip-active-record] # Skip Active Record files [--dev] # Setup the application with Gemfile pointing to your Rails checkout -J, [--skip-prototype] # Skip Prototype files -T, [--skip-test-unit] # Skip Test::Unit files -G, [--skip-git] # Skip Git ignores and keeps -r, [--ruby=PATH] # Path to the Ruby binary of your choice # Default: /System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/bin/ruby -m, [--template=TEMPLATE] # Path to an application template (can be a filesystem path or URL) -b, [--builder=BUILDER] # Path to an application builder (can be a filesystem path or URL) [--edge] # Setup the application with Gemfile pointing to Rails repository Runtime options: -q, [--quiet] # Supress status output -s, [--skip] # Skip files that already exist -f, [--force] # Overwrite files that already exist -p, [--pretend] # Run but do not make any changes Rails options: -h, [--help] # Show this help message and quit -v, [--version] # Show Rails version number and quit Description: The 'rails new' command creates a new Rails application with a default directory structure and configuration at the path you specify. Example: rails new ~/Code/Ruby/weblog This generates a skeletal Rails installation in ~/Code/Ruby/weblog. See the README in the newly created application to get going.

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  • Use localeURL middleware with apache prefix

    - by Olivier R.
    Good morning everyone, I Got a question about localeURL usage. Everything works great for me with url like this : www.mysite.com/ If I type www.mysite.com/ in adress bar, it turns correctly in www.mysite.com/en/ for example. If I use the view change_locale, it's also all right (ie change www.mysite.com/en/ in www.mysite.com/fr/). But my application use apache as server, and use a prefix for the site, that gives url like this : www.mysite.com/prefix/ If I type www.mysite.com/prefix/ in the adress bar, the adress turns into www.mysite.com/en/ without prefix (so 404) I change code of view to manage our settings.SERVER_PREFIX value : def change_locale(request) : """ Redirect to a given url while changing the locale in the path The url and the locale code need to be specified in the request parameters. O. Rochaix; Taken from localeURL view, and tuned to manage : - SERVER_PREFIX from settings.py """ next = request.REQUEST.get('next', None) if not next: next = request.META.get('HTTP_REFERER', None) if not next: next = settings.SERVER_PREFIX + '/' next = urlsplit(next).path prefix = False if settings.SERVER_PREFIX!="" and next.startswith(settings.SERVER_PREFIX) : prefix = True next = "/" + next.lstrip(settings.SERVER_PREFIX) _, path = utils.strip_path (next) if request.method == 'POST': locale = request.POST.get('locale', None) if locale and check_for_language(locale): path = utils.locale_path(path, locale) if prefix : path = settings.SERVER_PREFIX + path response = http.HttpResponseRedirect(path) return response with this customized view, i'm able to correctly change language, but i'm not sure that's the right way of doing stuff. Is there any option on localeURL to manage prefix of apache ?

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  • Matching Regular Expression in Javascript and PHP problem...

    - by Frankie
    I can't figure out how to get the same result from my Javscript as I do from my PHP. In particular, Javascript always leaves out the backslashes. Please ignore the random forward and backslashes; I put them there so that I can cover my basis on a windows system or any other system. Output: Input String: "/root\wp-cont ent\@*%'i@$@%$&^(@#@''mage6.jpg:" /root\wp-content\image6.jpg (PHP Output) /rootwp-contentimage6.jpg (Javscript Output) I would appreciate any help! PHP: <?php $path ="/root\wp-cont ent\@*%'i@$@%$&^(@#@''mage6.jpg:"; $path = preg_replace("/[^a-zA-Z0-9\\\\\/\.-]/", "", $path); echo $path; ?> Javascript: <script type="text/javascript"> var path = "/root\wp-cont ent\@*%'i@$@%$&^(@#@''mage6.jpg:"; //exact same string as PHP var regx = /[^a-zA-Z0-9\.\/-]/g; path = path.replace(regx,""); document.write("<br>"+path); </script>

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  • django inner redirects

    - by Zayatzz
    Hello I have one project that in my own development computer (uses mod_wsgi to serve the project) caused no problems. In live server (uses mod_fastcgi) it generates 500 though. my url conf is like this: # -*- coding: utf-8 -*- from django.conf.urls.defaults import * # Uncomment the next two lines to enable the admin: from django.contrib import admin admin.autodiscover() urlpatterns = patterns('', url(r'^admin/', include(admin.site.urls)), url(r'^', include('jalka.game.urls')), ) and # -*- coding: utf-8 -*- from django.conf.urls.defaults import * from django.contrib.auth import views as auth_views urlpatterns = patterns('jalka.game.views', url(r'^$', view = 'front', name = 'front',), url(r'^ennusta/(?P<game_id>\d+)/$', view = 'ennusta', name = 'ennusta',), url(r'^login/$', auth_views.login, {'template_name': 'game/login.html'}, name='auth_login'), url(r'^logout/$', auth_views.logout, {'template_name': 'game/logout.html'}, name='auth_logout'), url(r'^arvuta/$', view = 'arvuta', name = 'arvuta',), ) and .htaccess is like that: Options +FollowSymLinks RewriteEngine on RewriteOptions MaxRedirects=10 # RewriteCond %{HTTP_HOST} . RewriteCond %{HTTP_HOST} ^www\.domain\.com RewriteRule (.*) http://domain.com/$1 [R=301,L] AddHandler fastcgi-script .fcgi RewriteCond %{HTTP_HOST} ^jalka\.domain\.com$ [NC] RewriteCond %{REQUEST_FILENAME} !-f RewriteRule ^(.*) cgi-bin/fifa2010.fcgi/$1 [QSA,L] RewriteCond %{HTTP_HOST} ^subdomain\.otherdomain\.eu$ [NC] RewriteCond %{REQUEST_FILENAME} !-f RewriteRule ^(.*) cgi-bin/django.fcgi/$1 [QSA,L] Notice, that i have also other project set up with same .htaccess and that one is running just fine with more complex urls and views fifa2010.fcgi: #!/usr/local/bin/python # -*- coding: utf-8 -*- import sys, os DOMAIN = "domain.com" APPNAME = "jalka" PREFIX = "/www/apache/domains/www.%s" % (DOMAIN,) # Add a custom Python path. sys.path.insert(0, os.path.join(PREFIX, "htdocs/django/Django-1.2.1")) sys.path.insert(0, os.path.join(PREFIX, "htdocs")) sys.path.insert(0, os.path.join(PREFIX, "htdocs/jalka")) # Switch to the directory of your project. (Optional.) os.chdir(os.path.join(PREFIX, "htdocs", APPNAME)) # Set the DJANGO_SETTINGS_MODULE environment variable. os.environ['DJANGO_SETTINGS_MODULE'] = "%s.settings" % (APPNAME,) from django.core.servers.fastcgi import runfastcgi runfastcgi(method="threaded", daemonize="false") Alan

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  • WCF done broke.

    - by SteveCav
    WCF is completely fouled up on my machine. I start a new sln in VS2008. Add a project (WCF Service Lib).Build the project (builds ok). debug the project -- it gets halfway through the progress bar and bombs out with the error below (real path changed to "my mex path" so codeproject doesn't think its spam). I'm assuming it's file permissions but don't know where to start. Any ideas?: Error: Cannot obtain Metadata from my mex path If this is a Windows (R) Communication Foundation service to which you have access, please check that you have enabled metadata publishing at the specified address. For help enabling metadata publishing, please refer to the MSDN documentation at somewhere unhelpful Exchange Error URI: my mex path Metadata contains a reference that cannot be resolved: 'my mex path'. Could not connect to my mex path. TCP error code 10061: No connection could be made because the target machine actively refused it 127.0.0.1:8731. Unable to connect to the remote server No connection could be made because the target machine actively refused it 127.0.0.1:8731HTTP GET Error URI: my mex path There was an error downloading 'my mex path'. Unable to connect to the remote server No connection could be made because the target machine actively refused it 127.0.0.1:8731

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  • django+mod_wsgi on virtualenv not working

    - by jwesonga
    I've just finished setting up a django app on virtualenv, deployment went smoothly using a fabric script, but now the .wsgi is not working, I've tried every variation on the internet but no luck. My .wsgi file is: import os import sys import django.core.handlers.wsgi # put the Django project on sys.path root_path = os.path.abspath(os.path.dirname(__file__) + '../') sys.path.insert(0, os.path.join(root_path, 'kcdf')) sys.path.insert(0, root_path) os.environ['DJANGO_SETTINGS_MODULE'] = 'kcdf.settings' application = django.core.handlers.wsgi.WSGIHandler() I keep getting the same error: [Sun Apr 18 12:44:30 2010] [error] [client 41.215.123.159] mod_wsgi (pid=16938): Exception occurred processing WSGI script '/home/kcdfweb/webapps/kcdf.web/releases/current/kcdf/apache/kcdf.wsgi'. [Sun Apr 18 12:44:30 2010] [error] [client 41.215.123.159] Traceback (most recent call last): [Sun Apr 18 12:44:30 2010] [error] [client 41.215.123.159] File "/usr/local/lib/python2.6/dist-packages/django/core/handlers/wsgi.py", line 230, in __call__ [Sun Apr 18 12:44:30 2010] [error] [client 41.215.123.159] self.load_middleware() [Sun Apr 18 12:44:30 2010] [error] [client 41.215.123.159] File "/usr/local/lib/python2.6/dist-packages/django/core/handlers/base.py", line 33, in load_middleware [Sun Apr 18 12:44:30 2010] [error] [client 41.215.123.159] for middleware_path in settings.MIDDLEWARE_CLASSES: [Sun Apr 18 12:44:30 2010] [error] [client 41.215.123.159] File "/usr/local/lib/python2.6/dist-packages/django/utils/functional.py", line 269, in __getattr__ [Sun Apr 18 12:44:30 2010] [error] [client 41.215.123.159] self._setup() [Sun Apr 18 12:44:30 2010] [error] [client 41.215.123.159] File "/usr/local/lib/python2.6/dist-packages/django/conf/__init__.py", line 40, in _setup [Sun Apr 18 12:44:30 2010] [error] [client 41.215.123.159] self._wrapped = Settings(settings_module) [Sun Apr 18 12:44:30 2010] [error] [client 41.215.123.159] File "/usr/local/lib/python2.6/dist-packages/django/conf/__init__.py", line 75, in __init__ [Sun Apr 18 12:44:30 2010] [error] [client 41.215.123.159] raise ImportError, "Could not import settings '%s' (Is it on sys.path? Does it have syntax errors?): %s" % (self.SETTINGS_MODULE, e) [Sun Apr 18 12:44:30 2010] [error] [client 41.215.123.159] ImportError: Could not import settings 'kcdf.settings' (Is it on sys.path? Does it have syntax errors?): No module named kcdf.settings my virtual environment is on /home/user/webapps/kcdfweb my app is /home/user/webapps/kcdf.web/releases/current/project_name my wsgi file home/user/webapps/kcdf.web/releases/current/project_name/apache/project_name.wsgi

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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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  • VBA: How go I get the total width from all controls in an MS-Access form?

    - by Stefan Åstrand
    Hi, This is probably very basic stuff, but please bear in mind I am completely new to these things. I am working on a procedure for my Access datasheet forms that will: Adjust the width of each column to fit content Sum the total width of all columns and subtract it from the size of the window's width Adjust the width of one of the columns to fit the remaining space This is the code that adjusts the width of each column to fit content (which works fine): Dim Ctrl As Control Dim Path As String Dim ClmWidth As Integer 'Adjust column width to fit content For Each Ctrl In Me.Controls If TypeOf Ctrl Is TextBox Then Path = Ctrl.Name Me(Path).ColumnWidth = -2 End If Next Ctrl How should I write the code so I get the total width of all columns? Thanks a lot! Stefan Solution This is the code that makes an Access datasheet go from this: To this: Sub AdjustColumnWidth(frm As Form, clm As String) On Error GoTo HandleError Dim intWindowWidth As Integer ' Window width property Dim ctrl As Control ' Control Dim intCtrlWidth As Integer ' Control width property Dim intCtrlSum As Integer ' Control width property sum Dim intCtrlAdj As Integer ' Control width property remaining after substracted intCtrSum 'Adjust column width to standard width For Each ctrl In frm.Controls If TypeOf ctrl Is TextBox Or TypeOf ctrl Is CheckBox Or TypeOf ctrl Is ComboBox Then Path = ctrl.Name frm(Path).ColumnWidth = 1500 End If Next ctrl 'Get total column width For Each ctrl In frm.Controls If TypeOf ctrl Is TextBox Or TypeOf ctrl Is CheckBox Or TypeOf ctrl Is ComboBox Then Path = ctrl.Name intCtrlWidth = frm(Path).ColumnWidth If Path <> clm Then intCtrlSum = intCtrlSum + intCtrlWidth End If End If Next ctrl 'Adjust column to fit window intWindowWidth = frm.WindowWidth - 270 intCtrlAdj = intWindowWidth - intCtrlSum frm.Width = intWindowWidth frm(clm).ColumnWidth = intCtrlAdj Debug.Print "Totalt (Ctrl): " & intCtrlSum Debug.Print "Totalt (Window): " & intWindowWidth Debug.Print "Totalt (Remaining): " & intCtrlAdj Debug.Print "clm : " & clm HandleError: GeneralErrorHandler Err.Number, Err.Description Exit Sub End Sub Code to call procedure: Private Sub Form_Load() Call AdjustColumnWidth(Me, "txtDescription") End Sub

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  • How to create a semi transparent window in WPF that allows mouse events to pass through

    - by RMK
    I am trying to create an effect similar to the Lights out /lights dim feature in Adobe Lightroom (http://www.youtube.com/watch?v=87hNd3vaENE) except in WPF. What I tried was to create another window over-top of my existing window, make it transparent and put a semi transparent Path geometry on it. But I want mouse events to be able to pass through this semi transparent window (on to windows below). This is a simplified version of what I have: <Window x:Class="LightsOut.MaskWindow" xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation" xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml" AllowsTransparency="True" WindowStyle="None" ShowInTaskbar="False" Topmost="True" Background="Transparent"> <Grid> <Button HorizontalAlignment="Left" Height="20" Width="60">click</Button> <Path IsHitTestVisible="False" Stroke="Black" Fill="Black" Opacity="0.3"> <Path.Data> <RectangleGeometry Rect="0,0,1000,1000 "/> </Path.Data> </Path> </Grid> The window is fully transparent, so on places where the Path doesn't cover, mouse events pass right through. So far so good. The IsHitTestvisible is set to false on the path object. So mouse events will pass through it to other controls on the same form (ie you can click on the Button, because it is on the same form). But mouse events wont pass through the Path object onto windows that are below it. Any ideas? Or better ways to solve this problem? Thanks.

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  • Baffled by differences between WPF BitmapEncoders

    - by DanM
    I wrote a little utility class that saves BitmapSource objects to image files. The image files can be either bmp, jpeg, or png. Here is the code: public class BitmapProcessor { public void SaveAsBmp(BitmapSource bitmapSource, string path) { Save(bitmapSource, path, new BmpBitmapEncoder()); } public void SaveAsJpg(BitmapSource bitmapSource, string path) { Save(bitmapSource, path, new JpegBitmapEncoder()); } public void SaveAsPng(BitmapSource bitmapSource, string path) { Save(bitmapSource, path, new PngBitmapEncoder()); } private void Save(BitmapSource bitmapSource, string path, BitmapEncoder encoder) { using (var stream = new FileStream(path, FileMode.Create)) { encoder.Frames.Add(BitmapFrame.Create(bitmapSource)); encoder.Save(stream); } } } Each of the three Save methods work, but I get unexpected results with bmp and jpeg. Png is the only format that produces an exact reproduction of what I see if I show the BitmapSource on screen using a WPF Image control. Here are the results: BMP - too dark JPEG - too saturated PNG - correct Why am I getting completely different results for different file types? I should note that the BitmapSource in my example uses an alpha value of 0.1 (which is why it appears very desaturated), but it should be possible to show the resulting colors in any image format. I know if I take a screen capture using something like HyperSnap, it will look correct regardless of what file type I save to. Here's a HyperSnap screen capture saved as a bmp: As you can see, this isn't a problem, so there's definitely something strange about WPF's image encoders. Do I have a setting wrong? Am I missing something?

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  • HttpPostedFile.SaveAs() throws UnauthorizedAccessException even though the file is saved?

    - by jrummell
    I have an aspx page with multiple FileUpload controls and one Upload button. In the click handler I save the files like this: string path = "..."; for (int i = 0; i < Request.Files.Count - 1; i++) { HttpPostedFile file = Request.Files[i]; string fileName = Path.GetFileName(file.FileName); string saveAsPath = Path.Combine(path, fileName); file.SaveAs(saveAsPath); } When file.SaveAs() is called, it throws: System.Web.HttpUnhandledException: Exception of type 'System.Web.HttpUnhandledException' was thrown. --- System.UnauthorizedAccessException: Access to the path '...' is denied. at System.IO.__Error.WinIOError(Int32 errorCode, String maybeFullPath) at System.IO.FileStream.Init(String path, FileMode mode, FileAccess access, Int32 rights, Boolean useRights, FileShare share, Int32 bufferSize, FileOptions options, SECURITY_ATTRIBUTES secAttrs, String msgPath, Boolean bFromProxy) at System.IO.FileStream..ctor(String path, FileMode mode, FileAccess access, FileShare share, Int32 bufferSize, FileOptions options, String msgPath, Boolean bFromProxy) at System.IO.FileStream..ctor(String path, FileMode mode) at System.Web.HttpPostedFile.SaveAs(String filename) at Belden.Web.Intranet.Iso.Complaints.AttachmentUploader.btnUpload_Click(Object sender, EventArgs e) at System.Web.UI.WebControls.Button.OnClick(EventArgs e) at System.Web.UI.WebControls.Button.RaisePostBackEvent(String eventArgument) at System.Web.UI.Page.RaisePostBackEvent(IPostBackEventHandler sourceControl, String eventArgument) at System.Web.UI.Page.ProcessRequestMain(Boolean includeStagesBeforeAsyncPoint, Boolean includeStagesAfterAsyncPoint) --- End of inner exception stack trace --- at System.Web.UI.Page.HandleError(Exception e) at System.Web.UI.Page.ProcessRequestMain(Boolean includeStagesBeforeAsyncPoint, Boolean includeStagesAfterAsyncPoint) at System.Web.UI.Page.ProcessRequest(Boolean includeStagesBeforeAsyncPoint, Boolean includeStagesAfterAsyncPoint) at System.Web.UI.Page.ProcessRequest() at System.Web.UI.Page.ProcessRequest(HttpContext context) at ASP.departments_iso_complaints_uploadfiles_aspx.ProcessRequest(HttpContext context) at System.Web.HttpApplication.CallHandlerExecutionStep.System.Web.HttpApplication.IExecutionStep.Execute() at System.Web.HttpApplication.ExecuteStep(IExecutionStep step, Boolean& completedSynchronously) Now here's the fun part. The file is saved correctly! So why is it throwing this exception?

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  • More efficient approach to XSLT for-each

    - by Paul
    I have an XSLT which takes a . delimted string and splits it into two fields for a SQL statement: <xsl:for-each select="tokenize(Path,'\.')"> <xsl:choose> <xsl:when test="position() = 1 and position() = last()">SITE = '<xsl:value-of select="."/>' AND PATH = ''</xsl:when> <xsl:when test="position() = 1 and position() != last()">SITE = '<xsl:value-of select="."/>' </xsl:when> <xsl:when test="position() = 2 and position() = last()">AND PATH = '<xsl:value-of select="."/>' </xsl:when> <xsl:when test="position() = 2">AND PATH = '<xsl:value-of select="."/></xsl:when> <xsl:when test="position() > 2 and position() != last()">.<xsl:value-of select="."/></xsl:when> <xsl:when test="position() > 2 and position() = last()">.<xsl:value-of select="."/>' </xsl:when> <xsl:otherwise>zxyarglfaux</xsl:otherwise> </xsl:choose> </xsl:for-each> The results are as follows: INPUT: North OUTPUT: SITE = 'North' AND PATH = '' INPUT: North.A OUTPUT: SITE = 'North' AND PATH = 'A' INPUT: North.A.B OUTPUT: SITE = 'North' AND PATH = 'A.B' INPUT: North.A.B.C OUTPUT: SITE = 'North' AND PATH = 'A.B.C' This works, but is very lengthy. Can anyone see a more efficient approach? Thanks!

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  • Segmentation fault

    - by darkie15
    #include<stdio.h> #include<zlib.h> #include<unistd.h> #include<string.h> int main(int argc, char *argv[]) { char *path=NULL; size_t size; int index ; printf("\nArgument count is = %d", argc); printf ("\nThe 0th argument to the file is %s", argv[0]); path = getcwd(path, size); printf("\nThe current working directory is = %s", path); if (argc <= 1) { printf("\nUsage: ./output filename1 filename2 ..."); } else if (argc > 1) { for (index = 1; index <= argc;index++) { printf("\n File name entered is = %s", argv[index]); strcat(path,argv[index]); printf("\n The complete path of the file name is = %s", path); } } return 0; } In the above code, here is the output that I get while running the code: $ ./output test.txt Argument count is = 2 The 0th argument to the file is ./output The current working directory is = /home/welcomeuser File name entered is = test.txt The complete path of the file name is = /home/welcomeusertest.txt Segmentation fault (core dumped) Can anyone please me understand why I am getting a core dumped error? Regards, darkie

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  • Failure to connect to admin share pops up dialog

    - by Jan
    I'm having an issue with a curious error message when accessing the administrative share on a remote machine. Specifically, the client is logged in as the domain administrator on the machine A, and runs some code that tries to access the admin share on B (a domain member). The access is done in .NET, along these lines (though I am not sure if the method of access makes a difference): string path = @"\\B\admin$"; if (Directory.Exists(path)) { try { path += @"\temp\"; if (!Directory.Exists(path)) { Directory.CreateDirectory(path); } path += "myfile_remote"; File.Copy("myfile", path); Now, on some machines this fails. That is not a big problem as we have a fallback. I'd like to know why but it is not the real issue. The problem is that running this piece of code causes a dialog box to pop up for the logged-in user on B, saying "network error trying to access \\B\admin$\temp\myfile_remote. Contact the network administrator and ask for the correct permissions". Unfortunately, it is a foreign language Windows so I'll spare you all posting a screenshot. It is skinned like a standard Windows dialog box. Why exactly is that dialog box popping up for the user and is there anything I can do about it? Edit to add: B is a Windows 7 Enterprise installation. The client is not aware of any GPO policies being installed. There is AV from Trend Micro installed.

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