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  • Ruby GraphViz Binary Tree Record

    - by Jason M
    I'm using the ruby-graphviz gem and I'm trying to draw binary trees. I'd like to use the record shape so that each node can have a left, middle, and right field and, thus, if there are two edges leaving a node, the left and right edges can be distinguished. I tried specifying the field by concatenating the field name like this: @node1.name + ":left" But that did not work. What is the correct way of specifying the field? require 'rubygems' require 'graphviz' @graph = GraphViz.new( :G, :type => :digraph ) @node1 = @graph.add_node("1", "shape" => "record", "label" => "<left>|<f1> 1|<right>" ) @node2 = @graph.add_node("2", "shape" => "record", "label" => "<left>|<f1> 2|<right>" ) @graph.add_edge(@node1.name + ":left", @node2) # generate a random filename filename = "/tmp/#{(0...8).map{65.+(rand(25)).chr}.join}.png" @graph.output( :png => filename ) exec "open #{filename}"

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  • Adding a new target type to msbuild: How do I refer to the itemname in the task rules?

    - by jmucchiello
    I'm trying to add a task to build the COM proxy DLL after building the main DLL. So I created the following in a .target file: <Target Name="ProxyDLL" Inputs="$(IntDir)%(WHATGOESHERE)_i.c;$(IntDir)dlldata.c" Outputs="$(OutDir)%(WHATGOESHERE)ps.dll" AfterTargets="Link"> <CL Sources="$(IntDir)%(WHATGOESHERE)_i.c;$(IntDir)dlldata.c" /> </Target> And reference it from the .vcxproj file as <ItemGroup> <ProxyDLL Include="FTAccountant" /> </ItemGroup> So the FTAccountant.DLL file is created through the normal build process and then when attempts to compile the proxy stubs it creates these command lines: cl /c dir\_i.c dir\dlldata.c And of course it can't find _i.c. The first attempt, I put %(Filename) in the WHATGOESHERE space and I got this error: C:\ActivePay\Build\Proxy DLL.targets(6,3): error MSB4095: The item metadata %(Filename) is being referenced without an item name. Specify the item name by using %(itemname.Filename). So I changed it to %(itemname.Filename) and that is an empty string. How to get the value specified in the task's Include attribute and use it within the task?

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  • How can i zip files in Java and not include files paths

    - by Ignacio
    For example, i want to zip a file stored in /Users/me/Desktop/image.jpg I maded this method: public static Boolean generateZipFile(ArrayList<String> sourcesFilenames, String destinationDir, String zipFilename){ // Create a buffer for reading the files byte[] buf = new byte[1024]; try { // VER SI HAY QUE CREAR EL ROOT PATH boolean result = (new File(destinationDir)).mkdirs(); String zipFullFilename = destinationDir + "/" + zipFilename ; System.out.println(result); // Create the ZIP file ZipOutputStream out = new ZipOutputStream(new FileOutputStream(zipFullFilename)); // Compress the files for (String filename: sourcesFilenames) { FileInputStream in = new FileInputStream(filename); // Add ZIP entry to output stream. out.putNextEntry(new ZipEntry(filename)); // Transfer bytes from the file to the ZIP file int len; while ((len = in.read(buf)) > 0) { out.write(buf, 0, len); } // Complete the entry out.closeEntry(); in.close(); } // Complete the ZIP file out.close(); return true; } catch (IOException e) { return false; } } But when i extract the file, the unzipped files have the full path. I don't want the full path of each file in the zip i only want the filename. How can i made this?

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  • File Handler returning garbled files.

    - by forcripesake
    For the past 3 months my site has been using a PHP file handler in combination with htaccess. Users accessing the uploads folder of the site would be redirected to the handler as such: RewriteRule ^(.+)\.*$ downloader.php?f=%{REQUEST_FILENAME} [L] The purpose of the file handler is pseudo coded below, followed by actual code. //Check if file exists and user is downloading from uploads directory; True. //Check against a file type white list and set the mime type(); $ctype = mime type; header("Pragma: public"); // required header("Expires: 0"); header("Cache-Control: must-revalidate, post-check=0, pre-check=0"); header("Cache-Control: private",false); // required for certain browsers header("Content-Type: $ctype"); header("Content-Disposition: attachment; filename=\"".basename($filename)."\";" ); header("Content-Transfer-Encoding: binary"); header("Content-Length: ".filesize($filename)); readfile("$filename"); As of yesterday, the handler started returning garbled files, unreadable images, and had to be bypassed. I'm wondering what settings could have gone awry to cause this.

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  • Changing an image's ALT value with jQuery

    - by NightMICU
    Hi all, I have a modal form that changes the caption of a photo (paragraph under the image) and I am also trying to change the image's ALT attribute but cannot seem to. Here is the jQuery I am trying to make work $(".edit").click(function() { var parent = $(this).parents('.item'); var caption = $(parent).find('.labelCaption').html(); $("#photoCaption").val(caption); $("#editCaptionDialog").dialog({ width: 450, bgiframe: true, resizable: false, modal: true, title: 'Edit Caption', overlay: { backgroundColor: '#000', opacity: 0.5 }, buttons: { 'Edit': function() { var newCaption = $("#photoCaption").val(); $(parent).find(".labelCaption").html(newCaption); $(parent).find('img').attr('alt', newCaption); } } }); return false; }); And the HTML <li class="item ui-corner-all" id="photo<? echo $images['id'];?>"> <div> <a href="http://tapp-essexvfd.org/gallery/photos/<?php echo $images['filename'];?>.jpg" class="lightbox" title="<?php echo $images['caption'];?>"> <img src="http://tapp-essexvfd.org/gallery/photos/thumbs/<?php echo $images['filename'];?>.jpg" alt="<?php echo $images['caption'];?>" class="photo ui-corner-all"/></a><br/> <p><span class="labelCaption"><?php echo $images['caption'];?> </span></p> <p><a href="edit_photo.php?filename=<?php echo $images['filename'];?>" class="button2 edit ui-state-default ui-corner-all">Edit</a></p> </div> </li> The caption is changing like it should. Thanks

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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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  • Files under Program Files have a split personality

    - by regularfry
    I have a Ruby application I'm installing (along with a packaged ruby interpreter) under Program Files on Windows 7 with an NSIS-built installer. In order to debug it, I edited one of the files to add some debugging statements. After that, I uninstalled the package and ran a new version of the installer which includes a new copy of the edited file, without debugging statements. Now, I can't get the new copy to load into ruby. If I run type <filename> in cmd.exe, or open the file in Notepad.exe or Firefox, I see the new version. If I run ruby -e "puts File.read('<filename>')", or open the file in emacs, I see the old version. If, in Windows Explorer, I copy the file to a new filename, everything can see the new contents at that filename. If I delete the original file and rename the copy to replace the original, the split personality returns. This situation survives a reboot, so it's not a simple matter of a file being accidentally held open. What on earth is going on here? Is there some aspect of the install process that might be checkpointing the file in a way I can revert, or at least switch off while I'm debugging the installer?

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  • how to update div tag in javascript with data from model for onsubmit form asp.net mvc

    - by michael
    In my page i have a form tag which submits to server ,gets data and redirects to same page. problem is the the div tag which has the data from server is not getting updated. how to do that in javascript <% using (Html.BeginForm("Addfile", "uploadfile", FormMethod.Post, new { id = "uploadform", enctype = "multipart/form-data" })) { %> <input type="file" id="addedFile" name="addedFile" /><br /> <input type="submit" id="addfile" value="Addfile" /> <div id="MyGrid"> //data from the model(server side) filelist is not updating</div> what will be the form onsubmit javascript function to update the div tag with the data from the model. and my uploadfile controller get post methods are as [AcceptVerbs(HttpVerbs.Get)] public ActionResult Upload() { return View(); } [AcceptVerbs(HttpVerbs.Post)] public ActionResult AddFile(HttpPostedFileBase addedFile) { static List<string> fileList = new List<string>(); string filename = Path.GetFileName(addedFile.FileName); file.SaveAs(@"D:\Upload\" + filename); fileList.Add(filename); return("Upload",fileList); } thanks, michaela

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  • getSystemResourceAsStream() returns null

    - by Hitesh Solanki
    Hiii... I want to get the content of properties file into InputStream class object using getSystemResourceAsStream(). I have built the sample code. It works well using main() method,but when i deploy the project and run on the server, properties file path cannot obtained ... so inputstream object store null value. Sample code is here.. public class ReadPropertyFromFile { public static Logger logger = Logger.getLogger(ReadPropertyFromFile.class); public static String readProperty(String fileName, String propertyName) { String value = null; try { //fileName = "api.properties"; //propertyName = "api_loginid"; System.out.println("11111111...In the read proprty file....."); // ClassLoader loader = ClassLoader.getSystemClassLoader(); InputStream inStream = ClassLoader.getSystemResourceAsStream(fileName); System.out.println("In the read proprty file....."); System.out.println("File Name :" + fileName); System.out.println("instream = "+inStream); Properties prop = new Properties(); try { prop.load(inStream); value = prop.getProperty(propertyName); } catch (Exception e) { logger.warn("Error occured while reading property " + propertyName + " = ", e); return null; } } catch (Exception e) { System.out.println("Exception = " + e); } return value; } public static void main(String args[]) { System.out.println("prop value = " + ReadPropertyFromFile.readProperty("api.properties", "api_loginid")); } }

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  • Download dynamic file with GWT

    - by Maksim
    I have a GWT page where user enter data (start date, end date, etc.), then this data goes to the server via RPC call. On the server I want to generate Excel report with POI and let user save that file on their local machine. This is my test code to stream file back to the client but for some reason I think it does not know how to stream file to the client when I'm using RPC: public class ReportsServiceImpl extends RemoteServiceServlet implements ReportsService { public String myMethod(String s) { File f = new File("/excelTestFile.xls"); String filename = f.getName(); int length = 0; try { HttpServletResponse resp = getThreadLocalResponse(); ServletOutputStream op = resp.getOutputStream(); ServletContext context = getServletConfig().getServletContext(); resp.setContentType("application/octet-stream"); resp.setContentLength((int) f.length()); resp.setHeader("Content-Disposition", "attachment; filename*=\"utf-8''" + filename + ""); byte[] bbuf = new byte[1024]; DataInputStream in = new DataInputStream(new FileInputStream(f)); while ((in != null) && ((length = in.read(bbuf)) != -1)) { op.write(bbuf, 0, length); } in.close(); op.flush(); op.close(); } catch (Exception ex) { ex.printStackTrace(); } return "Server says: " + filename; } } I've red somewhere on internet that you can't do file stream with RPC and I have to use Servlet for that. Is there any example of how to use Servlet and how to call that servlet from ReportsServiceImpl. Do I really need to make a servlet or it is possible to stream it back with my RPC?

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  • Unit Testing Private Method in Resource Managing Class (C++)

    - by BillyONeal
    I previously asked this question under another name but deleted it because I didn't explain it very well. Let's say I have a class which manages a file. Let's say that this class treats the file as having a specific file format, and contains methods to perform operations on this file: class Foo { std::wstring fileName_; public: Foo(const std::wstring& fileName) : fileName_(fileName) { //Construct a Foo here. }; int getChecksum() { //Open the file and read some part of it //Long method to figure out what checksum it is. //Return the checksum. } }; Let's say I'd like to be able to unit test the part of this class that calculates the checksum. Unit testing the parts of the class that load in the file and such is impractical, because to test every part of the getChecksum() method I might need to construct 40 or 50 files! Now lets say I'd like to reuse the checksum method elsewhere in the class. I extract the method so that it now looks like this: class Foo { std::wstring fileName_; static int calculateChecksum(const std::vector<unsigned char> &fileBytes) { //Long method to figure out what checksum it is. } public: Foo(const std::wstring& fileName) : fileName_(fileName) { //Construct a Foo here. }; int getChecksum() { //Open the file and read some part of it return calculateChecksum( something ); } void modifyThisFileSomehow() { //Perform modification int newChecksum = calculateChecksum( something ); //Apply the newChecksum to the file } }; Now I'd like to unit test the calculateChecksum() method because it's easy to test and complicated, and I don't care about unit testing getChecksum() because it's simple and very difficult to test. But I can't test calculateChecksum() directly because it is private. Does anyone know of a solution to this problem?

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  • Issue while saving image using savefiledialog

    - by user1097772
    I'm using savefiledialog to save an image. Canvas is picturebox and the loaded image is bitmap. When I try to save it the file is created but somehow corrupted. Cause when I try againt load the image or show in different viewer it doesn't work - I mean the saved file is corrupted. There is an method for saving image. private void saveFileDialog1_FileOk(object sender, CancelEventArgs e) { System.IO.FileStream fs = (System.IO.FileStream)saveFileDialog1.OpenFile(); try { switch (saveFileDialog1.FilterIndex) { case 1: canvas.Image.Save(saveFileDialog1.FileName, System.Drawing.Imaging.ImageFormat.Bmp); break; case 2: canvas.Image.Save(saveFileDialog1.FileName, System.Drawing.Imaging.ImageFormat.Jpeg); break; case 3: canvas.Image.Save(saveFileDialog1.FileName, System.Drawing.Imaging.ImageFormat.Png); break; case 4: canvas.Image.Save(saveFileDialog1.FileName, System.Drawing.Imaging.ImageFormat.Tiff); break; } } catch (Exception ex) { System.Console.WriteLine("Exception " + ex); } I should also mention the property Filter. saveFileDialog1.Filter has value: bmp (*.bmp)|*.bmp|jpeg (*.jpeg)|*.jpeg|png (*.png)|*.png|tiff (*.tiff)|*.tiff

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  • Download dynaic file with GWT

    - by Maksim
    I have a GWT page where user enter data (start date, end date, etc.), then this data goes to the server via RPC call. On the server I want to generate Excel report with POI and let user save that file on their local machine. This is my test code to stream file back to the client but for some reason it does not know: public class ReportsServiceImpl extends RemoteServiceServlet implements ReportsService { public String myMethod(String s) { File f = new File("/excelTestFile.xls"); String filename = f.getName(); int length = 0; try { HttpServletResponse resp = getThreadLocalResponse(); ServletOutputStream op = resp.getOutputStream(); ServletContext context = getServletConfig().getServletContext(); resp.setContentType("application/octet-stream"); resp.setContentLength((int) f.length()); resp.setHeader("Content-Disposition", "attachment; filename*=\"utf-8''" + filename + ""); byte[] bbuf = new byte[1024]; DataInputStream in = new DataInputStream(new FileInputStream(f)); while ((in != null) && ((length = in.read(bbuf)) != -1)) { op.write(bbuf, 0, length); } in.close(); op.flush(); op.close(); } catch (Exception ex) { ex.printStackTrace(); } return "Server says: " + filename; } } I've red somewhere on internet that you can't do file stream with RPC and I have to use Servlet for that. Is there any example of how to use Servlet and how to call that servlet from ReportsServiceImpl

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  • jQuery ajax post of jpg image to .net webservice. Image results corrupted

    - by sosergio
    I have a phonegap jquery app that opens the camera and take a picture. I then POST this picture to a .net webservice, wich I've coded. I can't use phonegap FileTransfer because such isn't supported by Bada os, wich is a requirement. I believe I've successfully loaded the image from phonegap FileSystem API, I've attached it into an .ajax type:post, I've even received it from .net side, but when .net save the image into the server, the image results corrupted. It seems to me that two sides of the communication have different data type. Has anyone experience in this? Any help will be appreciated. This is my code: //PHONEGAP CAMERA ACCESS (summed up) navigator.camera.getPicture(onGetPictureSuccess, onGetPictureFail, { quality: 50, destinationType:Camera.DestinationType.FILE_URI }); window.resolveLocalFileSystemURI(imageURI, onResolveFileSystemURISuccess, onResolveFileSystemURIError); fileEntry.file(gotFileSuccess, gotFileError); new FileReader().readAsDataURL(file); //UPLOAD FILE function onDataReadSuccess(evt) { var image_data = evt.target.result; var filename = unique_id(); var filext = "jpg"; $.ajax({ type : 'POST', url : SERVICE_BASE_URL+"/fotos/"+filename+"?ext="+filext, cache: false, timeout: 100000, processData: false, data: image_data, contentType: 'image/jpeg', success : function(data) { console.log("Data Uploaded with success. Message: "+ data); $.mobile.hidePageLoadingMsg(); $.mobile.changePage("ok.html"); } }); } On my .net Web Service this is the method that gets invoked: public string FotoSave(string filename, string extension, Stream fileContent) { string filePath = HttpContext.Current.Server.MapPath("~/foto_data/") + "\\" + filename; FileStream writeStream = new FileStream(filePath, FileMode.OpenOrCreate, FileAccess.Write); int Length = 256; Byte[] buffer = new Byte[Length]; int bytesRead = readStream.Read(buffer, 0, Length); // write the required bytes while (bytesRead > 0) { writeStream.Write(buffer, 0, bytesRead); bytesRead = readStream.Read(buffer, 0, Length); } readStream.Close(); writeStream.Close(); }

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  • git-diff to ignore ^M

    - by neoneye
    In a project where some of the files contains ^M as newline separators. Diffing these files are apparently impossible, since git-diff sees it as the entire file is just a single line. How does one diff with the previous version? Is there an option like "treat ^M as newline when diffing" ? prompt> git-diff "HEAD^" -- MyFile.as diff --git a/myproject/MyFile.as b/myproject/MyFile.as index be78321..a393ba3 100644 --- a/myproject/MyFile.cpp +++ b/myproject/MyFile.cpp @@ -1 +1 @@ -<U+FEFF>import flash.events.MouseEvent;^Mimport mx.controls.*;^Mimport mx.utils.Delegate \ No newline at end of file +<U+FEFF>import flash.events.MouseEvent;^Mimport mx.controls.*;^Mimport mx.utils.Delegate \ No newline at end of file prompt> UPDATE: now I have written a script that checks out the latest 10 revisions and converts CR to LF. require 'fileutils' if ARGV.size != 3 puts "a git-path must be provided" puts "a filename must be provided" puts "a result-dir must be provided" puts "example:" puts "ruby gitcrdiff.rb project/dir1/dir2/dir3/ SomeFile.cpp tmp_somefile" exit(1) end gitpath = ARGV[0] filename = ARGV[1] resultdir = ARGV[2] unless FileTest.exist?(".git") puts "this command must be run in the same dir as where .git resides" exit(1) end if FileTest.exist?(resultdir) puts "the result dir must not exist" exit(1) end FileUtils.mkdir(resultdir) 10.times do |i| revision = "^" * i cmd = "git show HEAD#{revision}:#{gitpath}#{filename} | tr '\\r' '\\n' > #{resultdir}/#{filename}_rev#{i}" puts cmd system cmd end

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  • Why is calling close() after fopen() not closing?

    - by Richard Morgan
    I ran across the following code in one of our in-house dlls and I am trying to understand the behavior it was showing: long GetFD(long* fd, const char* fileName, const char* mode) { string fileMode; if (strlen(mode) == 0 || tolower(mode[0]) == 'w' || tolower(mode[0]) == 'o') fileMode = string("w"); else if (tolower(mode[0]) == 'a') fileMode = string("a"); else if (tolower(mode[0]) == 'r') fileMode = string("r"); else return -1; FILE* ofp; ofp = fopen(fileName, fileMode.c_str()); if (! ofp) return -1; *fd = (long)_fileno(ofp); if (*fd < 0) return -1; return 0; } long CloseFD(long fd) { close((int)fd); return 0; } After repeated calling of GetFD with the appropriate CloseFD, the whole dll would no longer be able to do any file IO. I wrote a tester program and found that I could GetFD 509 times, but the 510th time would error. Using Process Explorer, the number of Handles did not increase. So it seems that the dll is reaching the limit for the number of open files; setting _setmaxstdio(2048) does increase the amount of times we can call GetFD. Obviously, the close() is working quite right. After a bit of searching, I replaced the fopen() call with: long GetFD(long* fd, const char* fileName, const char* mode) { *fd = (long)open(fileName, 2); if (*fd < 0) return -1; return 0; } Now, repeatedly calling GetFD/CloseFD works. What is going on here?

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  • Saving animated GIFs using urllib.urlopen (image saved does not animate)

    - by wenbert
    I have Apache2 + Django + X-sendfile. My problem is that when I upload an animated GIF, it won't "animate" when I output through the browser. Here is my code to display the image located outside the public accessible directory. def raw(request,uuid): target = str(uuid).split('.')[:-1][0] image = Uploads.objects.get(uuid=target) path = image.path filepath = os.path.join(path,"%s.%s" % (image.uuid,image.ext)) response = HttpResponse(mimetype=mimetypes.guess_type(filepath)) response['Content-Disposition']='filename="%s"'\ %smart_str(image.filename) response["X-Sendfile"] = filepath response['Content-length'] = os.stat(filepath).st_size return response UPDATE It turns out that it works. My problem is when I try to upload an image via URL. It probably doesn't save the entire GIF? def handle_url_file(request): """ Open a file from a URL. Split the file to get the filename and extension. Generate a random uuid using rand1() Then save the file. Return the UUID when successful. """ try: file = urllib.urlopen(request.POST['url']) randname = rand1(settings.RANDOM_ID_LENGTH) newfilename = request.POST['url'].split('/')[-1] ext = str(newfilename.split('.')[-1]).lower() im = cStringIO.StringIO(file.read()) # constructs a StringIO holding the image img = Image.open(im) filehash = checkhash(im) image = Uploads.objects.get(filehash=filehash) uuid = image.uuid return "%s" % (uuid) except Uploads.DoesNotExist: img.save(os.path.join(settings.UPLOAD_DIRECTORY,(("%s.%s")%(randname,ext)))) del img filesize = os.stat(os.path.join(settings.UPLOAD_DIRECTORY,(("%s.%s")%(randname,ext)))).st_size upload = Uploads( ip = request.META['REMOTE_ADDR'], filename = newfilename, uuid = randname, ext = ext, path = settings.UPLOAD_DIRECTORY, views = 1, bandwidth = filesize, source = request.POST['url'], size = filesize, filehash = filehash, ) upload.save() #return uuid return "%s" % (upload.uuid) except IOError, e: raise e Any ideas? Thanks! Wenbert

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  • "A generic error occurred in GDI+" error while showing uploaded images

    - by Prasad
    i am using the following code to show the image that has been saved in my database from my asp.net mvc(C#) application:. public ActionResult GetSiteHeaderLogo() { SiteHeader _siteHeader = new SiteHeader(); Image imgImage = null; long userId = Utility.GetUserIdFromSession(); if (userId > 0) { _siteHeader = this.siteBLL.GetSiteHeaderLogo(userId); if (_siteHeader.Logo != null && _siteHeader.Logo.Length > 0) { byte[] _imageBytes = _siteHeader.Logo; if (_imageBytes != null) { using (System.IO.MemoryStream imageStream = new System.IO.MemoryStream(_imageBytes)) { imgImage = Image.FromStream(imageStream); } } string sFileExtension = _siteHeader.FileName.Substring(_siteHeader.FileName.IndexOf('.') + 1, _siteHeader.FileName.Length - (_siteHeader.FileName.IndexOf('.') + 1)); Response.ContentType = Utility.GetContentTypeByExtension(sFileExtension.ToLower()); Response.Cache.SetCacheability(HttpCacheability.NoCache); Response.BufferOutput = false; if (imgImage != null) { ImageFormat _imageFormat = Utility.GetImageFormat(sFileExtension.ToLower()); imgImage.Save(Response.OutputStream, _imageFormat); imgImage.Dispose(); } } } return new EmptyResult(); } It works fine when i upload original image. But when i upload any downloaded images, it throws the following error: System.Runtime.InteropServices.ExternalException: A generic error occurred in GDI+. System.Runtime.InteropServices.ExternalException: A generic error occurred in GDI+. at System.Drawing.Image.Save(Stream stream, ImageCodecInfo encoder, EncoderParameters encoderParams) at System.Drawing.Image.Save(Stream stream, ImageFormat format) For. Ex: When i upload the original image, it shows as logo in my site and i downloaded that logo from the site and when i re-upload the same downloaded image, it throws the above error. It seems very weird to me and not able to find why its happening. Any ideas on this?

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  • ASP.NET MVC 2.0 + Implementation of a IRouteHandler goes not fire

    - by Peter
    Can anybody please help me with this as I have no idea why public IHttpHandler GetHttpHandler(RequestContext requestContext) is not executing. In my Global.asax.cs I have public class MvcApplication : System.Web.HttpApplication { public static void RegisterRoutes(RouteCollection routes) { routes.IgnoreRoute("{resource}.axd/{*pathInfo}"); routes.MapRoute( "Default", // Route name "{controller}/{action}/{id}", // URL with parameters new { controller = "Home", action = "Index", id = "" } // Parameter defaults ); routes.Add("ImageRoutes", new Route("Images/{filename}", new CustomRouteHandler())); } protected void Application_Start() { RegisterRoutes(RouteTable.Routes); } } //CustomRouteHandler implementation is below public class CustomRouteHandler : IRouteHandler { public IHttpHandler GetHttpHandler(RequestContext requestContext) { // IF I SET A BREAK POINT HERE IT DOES NOT HIT FOR SOME REASON. string filename = requestContext.RouteData.Values["filename"] as string; if (string.IsNullOrEmpty(filename)) { // return a 404 HttpHandler here } else { requestContext.HttpContext.Response.Clear(); requestContext.HttpContext.Response.ContentType = GetContentType(requestContext.HttpContext.Request.Url.ToString()); // find physical path to image here. string filepath = requestContext.HttpContext.Server.MapPath("~/logo.jpg"); requestContext.HttpContext.Response.WriteFile(filepath); requestContext.HttpContext.Response.End(); } return null; } } Can any body tell me what I'm missing here. Simply public IHttpHandler GetHttpHandler(RequestContext requestContext) does not fire. I havn't change anything in the web.config either. What I'm missing here? Please help.

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  • How should I return different types in a method based on the value of a string in Java?

    - by Siracuse
    I'm new to Java and I have come to having the following problem: I have created several classes which all implement the interface "Parser". I have a JavaParser, PythonParser, CParser and finally a TextParser. I'm trying to write a method so it will take either a File or a String (representing a filename) and return the appropriate parser given the extension of the file. Here is some psuedo-code of what I'm basically attempting to do: public Parser getParser(String filename) { String extension = filename.substring(filename.lastIndexOf(".")); switch(extension) { case "py": return new PythonParser(); case "java": return new JavaParser(); case "c": return new CParser(); default: return new TextParser(); } } In general, is this the right way to handle this situation? Also, how should I handle the fact that Java doesn't allow switching on strings? Should I use the .hashcode() value of the strings? I feel like there is some design pattern or something for handling this but it eludes me. Is this how you would do it?

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  • PHP Outputting File Attachments with Headers

    - by OneNerd
    After reading a few posts here I formulated this function which is sort of a mishmash of a bunch of others: function outputFile( $filePath, $fileName, $mimeType = '' ) { // Setup $mimeTypes = array( 'pdf' => 'application/pdf', 'txt' => 'text/plain', 'html' => 'text/html', 'exe' => 'application/octet-stream', 'zip' => 'application/zip', 'doc' => 'application/msword', 'xls' => 'application/vnd.ms-excel', 'ppt' => 'application/vnd.ms-powerpoint', 'gif' => 'image/gif', 'png' => 'image/png', 'jpeg' => 'image/jpg', 'jpg' => 'image/jpg', 'php' => 'text/plain' ); // Send Headers //-- next line fixed as per suggestion -- header('Content-Type: ' . $mimeTypes[$mimeType]); header('Content-Disposition: attachment; filename="' . $fileName . '"'); header('Content-Transfer-Encoding: binary'); header('Accept-Ranges: bytes'); header('Cache-Control: private'); header('Pragma: private'); header('Expires: Mon, 26 Jul 1997 05:00:00 GMT'); readfile($filePath); } I have a php page (file.php) which does something like this (lots of other code stripped out): // I run this thru a safe function not shown here $safe_filename = $_GET['filename']; outputFile ( "/the/file/path/{$safe_filename}", $safe_filename, substr($safe_filename, -3) ); Seems like it should work, and it almost does, but I am having the following issues: When its a text file, I am getting a strange symbol as the first letter in the text document When its a word doc, it is corrupt (presumably that same first bit or byte throwing things off). I presume all other file types will be corrupt - have not even tried them Any ideas on what I am doing wrong? Thanks - UPDATE: changed line of code as suggested - still same issue.

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  • PHP Include and sort by variable within file

    - by Jason Hoax
    I have written this PHP include-script but now I'm trying to sort the included files out by variables WITHIN the included php's. In other words, in each included PHP file there is a rating, now I want the ratings to be read so that when they are included they will be sorted out from highest to lowest. (scores are like 6.0 to 9.0) Kind Regards! $location = 'experiments/visualizations'; foreach (glob("$location/*.php") as $filename) { include $filename; } The included files are named randomly like: File1: $filename = "AAAA"; $projecttitle = "Project Name"; $description = "This totally explains the product"; $score = "7.6"; File 2: $filename = "BBBB"; $projecttitle = "Project Name2" $description = "This totally explains the product"; $score = "9.6"; As you can see 9.6 is higher than 7.6 but PHP sorts the includes out by name instead of variables within the file. I tried sorting, but I can't get it fixed. Help!

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  • how to change type of value in an php array and sorting it..is it possible ?

    - by justjoe
    hi, i got problem with my code and hopefully someone able to figure it out. The main purpose is to sort array based on its value (then reindex its numerical key). i got this sample of filename : $filename = array("index 198.php", "index 192.php", "index 144.php", "index 2.php", "index 1.php", "index 100.php", "index 111.php"); $alloutput = array(); //all of index in array foreach ($filename as $name) { preg_match('#(\d+)#', $name, $output); // take only the numerical from file name array_shift($output); // cleaned. the last code create duplicate numerical in $output, if (is_array($hasilku)) { $alloutput = array_merge($alloutput, $output); } } //try to check the type of every value in array foreach ($alloutput as $output) { if (is_array($hasil)) { echo "array true </br>"; } elseif (is_int($hasil)) { echo "integer true </br>"; } elseif (is_string($hasil)) { //the numerical taken from filename always resuld "string". echo "string true </br>"; } } the output of this code will be : Array ( [0] = 198 [1] = 192 [2] = 144 [3] = 2 [4] = 1 [5] = 100 [6] = 111 ) i have test every output in array. It's all string (and not numerical), So the question is how to change this string to integer, so i can sort it from the lowest into the highest number ? the main purpose of this code is how to output array where it had been sort from lowest to highest ?

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  • How can i get a file from remote machine?

    - by programmerist
    How can i get a file from remote computer? i know remote computer ip and 51124 port is open. i need this algorith: 1) Connect 192.xxx.x.xxx ip via 51124 port 2) filename:123456 (i want to search it on remote machine) 3) Get File 4) Save C:\ 51124 port is open. can i access and can i search any file according to filename? My code is below: IPEndPoint ipEnd = new IPEndPoint(IPAddress.Any, 51124); Socket sock = new Socket(AddressFamily.InterNetwork, SocketType.Stream, ProtocolType.IP); sock.Bind(ipEnd); sock.Listen(maxConnections); Socket serverSocket = sock.Accept(); byte[] data = new byte[bufferSize]; int received = serverSocket.Receive(data); int filenameLength = BitConverter.ToInt32(data, 0); string filename = Encoding.ASCII.GetString(data, 4, filenameLength); BinaryWriter bWrite = new BinaryWriter(File.Open(outPath + filename, FileMode.Create)); bWrite.Write(data, filenameLength + 4, received - filenameLength - 4); int received2 = serverSocket.Receive(data); while (received2 0) { bWrite.Write(data, 0, received2); received2 = serverSocket.Receive(data); } bWrite.Close(); serverSocket.Close(); sock.Close();

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  • Access denied to path , when uploading image to folder in server

    - by apekshabs
    Am getting error when you are going to upload the file on specified folder in the server. Here I am going to upload P6100083.jpg in storeimg folder. When I am going to upload I am getting the following error: Access to the path 'C:\inetpub\vhosts\bookmygroups.com\httpdocs\storeimg\P6100083.jpg' is denied. Can anyone help me... How to use permisiion and were to use... My code is while uploading image if (FileUpload1.HasFile) { float fileSize = FileUpload1.PostedFile.ContentLength; float floatConverttoKB = fileSize / 1024; float floatConverttoMB = floatConverttoKB / 1024; string DirName = "storeimg"; string savepath = Server.MapPath(DirName + "/"); DirectoryInfo dir = new DirectoryInfo(savepath); // string savepath = "C:\\Documents and Settings\\ssis3\\My Documents\\Visual Studio 2005\\WebSites\\finalbookgroups\\" + DirName + "\\"; if (fileSize < 4194304) { string filename = Server.HtmlEncode(FileUpload1.FileName); string extension = System.IO.Path.GetExtension(filename).ToUpper(); if (extension.Equals(".jpg") || extension.Equals(".JPG") || extension.Equals(".JPEG") || extension.Equals(".GIF")) { savepath += filename; FileUpload1.SaveAs(savepath); } } Thanks in advance

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