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  • Path to background in servlet

    - by kapil chhattani
    //the below line is the element of my HTML form which renders the image sent by the servlet written further below. <img style="margin-left:91px; margin-top:-6px;" class="image" src="http://www.abcd.com/captchaServlet"> I generate a captcha code using the following code in java. public class captchaServlet extends HttpServlet { protected void processRequest(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { int width = 150; int height = 50; int charsToPrint = 6; String elegibleChars = "ABCDEFGHJKLMPQRSTUVWXYabcdefhjkmnpqrstuvwxy1234567890"; char[] chars = elegibleChars.toCharArray(); StringBuffer finalString = new StringBuffer(); for ( int i = 0; i < charsToPrint; i++ ) { double randomValue = Math.random(); int randomIndex = (int) Math.round(randomValue * (chars.length - 1)); char characterToShow = chars[randomIndex]; finalString.append(characterToShow); } System.out.println(finalString); BufferedImage bufferedImage = new BufferedImage(width, height, BufferedImage.TYPE_INT_RGB); Graphics2D g2d = bufferedImage.createGraphics(); Font font = new Font("Georgia", Font.BOLD, 18); g2d.setFont(font); RenderingHints rh = new RenderingHints( RenderingHints.KEY_ANTIALIASING, RenderingHints.VALUE_ANTIALIAS_ON); rh.put(RenderingHints.KEY_RENDERING, RenderingHints.VALUE_RENDER_QUALITY); g2d.setRenderingHints(rh); GradientPaint gp = new GradientPaint(0, 0, Color.BLUE, 0, height/2, Color.black, true); g2d.setPaint(gp); g2d.fillRect(0, 0, width, height); g2d.setColor(new Color(255, 255, 0)); Random r = new Random(); int index = Math.abs(r.nextInt()) % 5; char[] data=new String(finalString).toCharArray(); String captcha = String.copyValueOf(data); int x = 0; int y = 0; for (int i=0; i<data.length; i++) { x += 10 + (Math.abs(r.nextInt()) % 15); y = 20 + Math.abs(r.nextInt()) % 20; g2d.drawChars(data, i, 1, x, y); } g2d.dispose(); response.setContentType("image/png"); OutputStream os = response.getOutputStream(); ImageIO.write(bufferedImage, "png", os); os.close(); } protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { processRequest(request, response); } protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { processRequest(request, response); } } But in the above code background is also generated using the setPaint menthod I am guessing. I want the background to be some image from my local machine whoz URL i should be able to mention like URL url=this.getClass().getResource("Desktop/images.jpg"); BufferedImage bufferedImage = ImageIO.read(url); I am just writing the above two lines for making the reader understand better what the issue is. Dont want to use the exact same commands. All I want is the the background of the captcha code generated should be an image of my choice.

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  • htaccess redirect to subfolder with same path attached?

    - by Josh
    How can I go about redirecting my old URLs which would have been: http:// blah.com/some/post/name to the new URLs which would be: http:// blah.com/new/some/post/name Is this even possible? I don't want to simply redirect any requests for the blah.com domain to blah.com/new I want to make sure the subpath is still attached to the redirect

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  • please give me a solution

    - by user327832
    here is the code i have written so far but ended up giving me error import java.io.File; import java.io.FileInputStream; import java.io.IOException; import java.io.InputStream; public class Main { public static void main(String[] args) throws Exception { File file = new File("c:\\filea.txt"); InputStream is = new FileInputStream(file); long length = file.length(); System.out.println (length); bytes[] bytes = new bytes[(int) length]; try { int offset = 0; int numRead = 0; while (numRead >= 0) { numRead = is.read(bytes); } } catch (IOException e) { System.out.println ("Could not completely read file " + file.getName()); } is.close(); Object[] see = new Object[(int) length]; see[1] = bytes; System.out.println ((String[])see[1]); } }

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  • PHP: Fastest way possible to read contents of a file.

    - by SoLoGHoST
    Ok, I'm looking for the fastest possible way to read all of the contents of a file via php with a filepath on the server, also these files can be huge. So it's very important that it does a READ ONLY to it as fast as possible. Is reading it line by line faster than reading the entire contents? Though, I remember reading up on this some, that reading the entire contents can produce errors for huge files. Is this true?

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  • Winform .Net 3.5 - Get path of ApplicationFolder for Office Add-In

    - by Serexx
    Greetings - I am writing an add-in for Expression Web 3 in VS2008 The solution has a Windows Installer setup project in which the ApplicationFolder is tagged to hold some ancillary files. I need to be able to access those files during the add-in's startup, but naturally the Application object as well as the AppDomain reference paths to Expression Web not to the Add-In. There is a tickle in the back of my head that the solution involves reflection but thats as far as I can get... has anyone dealt with this or have any suggestions? Thanks!

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  • Anyone know any good backend user online file manager?

    - by skyhigh
    Hi I'm looking for a backend system where your clients can login and upload files to your server, download files from the server and you can delete the users, create users, etc. I do not know the proper name for this kind of software. Maybe its called online file manager? Any recommendations? My server supports PHP, apache and mysq. Thanks

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  • How can I get a filename from a path with Perl?

    - by Eric
    I am trying to parse the filename from paths. I have this: my $filepath = "/Users/Eric/Documents/foldername/filename.pdf"; $filepath =~ m/^.*\\(.*[.].*)$/; print "Linux path:"; print $1 . "\n\n"; print "-------\n"; my $filepath = "c:\\Windows\eric\filename.pdf"; $filepath =~ m/^.*\\(.*[.].*)$/; print "Windows path:"; print $1 . "\n\n"; print "-------\n"; my $filepath = "filename.pdf"; $filepath =~ m/^.*\\(.*[.].*)$/; print "Without path:"; print $1 . "\n\n"; print "-------\n"; But that returns: Linux path: ------- Windows path:Windowsic ilename.pdf ------- Without path:Windowsic ilename.pdf ------- I am expecting this: Linux path: filename.pdf ------- Windows path: filename.pdf ------- Without path: filename.pdf ------- Can somebody please point out what I am doing wrong? Thanks! :)

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  • time on files differ by 1 sec. FAIL Robocopy sync

    - by csmba
    I am trying to use Robocopy to sync (/IMG) a folder on my PC and a shared network drive. The problem is that the file attributes differ by 1 sec on both locations (creation,modified and access). So every time I run robocopy, it syncs the file again... BTW, problem is the same if I delete the target file and robocopy it from new... still, new file has 1 sec different properties. Env Details: Source: Win 7 64 bit Target: WD My Book World Edition NAS 1TB which takes its time from online NTP pool.ntp.org (I don't know if file system is FAT or not)

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  • How to animate a path like it's being drawn, dot by dot? (Raphael.js)

    - by Anton
    How to animate a vector path like it's being drawn by hand? In other words, slowly show the path pixel by pixel. I'm using Raphaël.js, but if your answer is not library specific—like maybe there's some general programming pattern for doing that kind of thing (I'm fairly new to vector animation)—it's welcome! Update I know it's easy to do with straight paths, as easy as an example on that page:: path("M114 253").animate({path: "M114 253 L 234 253"}); But try to change code on that page, say, this way:: path("M114 26").animate({path: "M114 26 C 24 23 234 253 234 253"}); And you'll see what I mean. Path is certainly animated from it initial state (point "M114 26") to the end state (curve "C 24 23 234 253 234 253" starting on point "M114 26"), but not in a way specified in question, not like it's being drawn. (Sorry for not making clear from the start that I don't mean animating straight lines.) I don't see how animateAlong can do that either. It can animate an object along a path, and how can I make this path to gradually show itself while object is being animated along it?

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  • Is there a way to efficiently yield every file in a directory containing millions of files?

    - by Josh Smeaton
    I'm aware of os.listdir, but as far as I can gather, that gets all the filenames in a directory into memory, and then returns the list. What I want, is a way to yield a filename, work on it, and then yield the next one, without reading them all into memory. Is there any way to do this? I worry about the case where filenames change, new files are added, and files are deleted using such a method. Some iterators prevent you from modifying the collection during iteration, essentially by taking a snapshot of the state of the collection at the beginning, and comparing that state on each move operation. If there is an iterator capable of yielding filenames from a path, does it raise an error if there are filesystem changes (add, remove, rename files within the iterated directory) which modify the collection? There could potentially be a few cases that could cause the iterator to fail, and it all depends on how the iterator maintains state. Using S.Lotts example: filea.txt fileb.txt filec.txt Iterator yields filea.txt. During processing, filea.txt is renamed to filey.txt and fileb.txt is renamed to filez.txt. When the iterator attempts to get the next file, if it were to use the filename filea.txt to find it's current position in order to find the next file and filea.txt is not there, what would happen? It may not be able to recover it's position in the collection. Similarly, if the iterator were to fetch fileb.txt when yielding filea.txt, it could look up the position of fileb.txt, fail, and produce an error. If the iterator instead was able to somehow maintain an index dir.get_file(0), then maintaining positional state would not be affected, but some files could be missed, as their indexes could be moved to an index 'behind' the iterator. This is all theoretical of course, since there appears to be no built-in (python) way of iterating over the files in a directory. There are some great answers below, however, that solve the problem by using queues and notifications. Edit: The OS of concern is Redhat. My use case is this: Process A is continuously writing files to a storage location. Process B (the one I'm writing), will be iterating over these files, doing some processing based on the filename, and moving the files to another location. Edit: Definition of valid: Adjective 1. Well grounded or justifiable, pertinent. (Sorry S.Lott, I couldn't resist). I've edited the paragraph in question above.

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  • Changing image domain / path in css for production?

    - by Neil
    Currently, for things like background images, our css files have no domain specified. This works both in our development and production environments. background-image: url(/images/bg.png); For performance reasons (cookie-less domain), we'd like to switch this: background-image: url(http://staticimagedomain.com/images/bg.png); Ideally, we don't hard code those, so our development environments can still pull locally. Any thoughts on how to best achieve this?

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  • Need to have JProgress bar to measure progress when copying directories and files

    - by user1815823
    I have the below code to copy directories and files but not sure where to measure the progress. Can someone help as to where can I measure how much has been copied and show it in the JProgress bar public static void copy(File src, File dest) throws IOException{ if(src.isDirectory()){ if(!dest.exists()){ //checking whether destination directory exisits dest.mkdir(); System.out.println("Directory copied from " + src + " to " + dest); } String files[] = src.list(); for (String file : files) { File srcFile = new File(src, file); File destFile = new File(dest, file); copyFolder(srcFile,destFile); } }else{ InputStream in = new FileInputStream(src); OutputStream out = new FileOutputStream(dest); byte[] buffer = new byte[1024]; int length; while ((length = in.read(buffer)) > 0){ out.write(buffer, 0, length); } in.close(); out.close(); System.out.println("File copied from " + src + " to " + dest); }

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  • Referencing java resource files for cold fusion

    - by Chimeara
    I am using a .Jar file containing a .properties file in my CF code, however it seems unable to find the .properties file when run from CF. My java code is: String key =""; String value =""; try { File file = new File("src/test.properties"); FileInputStream fileInput = new FileInputStream(file); Properties properties = new Properties(); properties.load(fileInput); fileInput.close(); Enumeration enuKeys = properties.keys(); while (enuKeys.hasMoreElements()) { key = (String) enuKeys.nextElement(); value = properties.getProperty(key); //System.out.println(key + ": " + value); } } catch (FileNotFoundException e) { e.printStackTrace(); key ="error"; } catch (IOException e) { e.printStackTrace(); key ="error"; } return(key + ": " + value); I have my test.properties file in the project src folder, and make sure it is selected when compiling, when run from eclipse it gives the expected key and value, however when run from CF I get the caught errors. My CF code is simply: propTest = CreateObject("java","package.class"); testResults = propTest.main2(); Is there a special way to reference the .properties file so CF can access it, or do I need to include the file outside the .jar somewhere?

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  • how to give image path in java script in asp.net mvc

    - by Renu123
    i am using asp.net mvc for my project. i am using partial views for that i used ajax jaquery every thing is working fine but i am not getting image for loading when i upload an image. image get displyed at local host it has problem when i upload the project. i have given image as: $jq("#ajaxThrobber").html('<img src="/Images/ajax-loader.gif"> Loading.....'); please advice me thank you.

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  • Specifying Entire Path As Optional Rails 3.0.0

    - by Kevin Sylvestre
    I want to create a Rails 3 route with entirely optional parameters. The example route is: match '(/name/:name)(/height/:height)(/weight/:weight)' => 'people#index' The route works if I specify it as: match '/people(/name/:name)(/height/:height)(/weight/:weight)' => 'people#index' But I want to have this as the root URL. Thanks.

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  • Get DLL's directory

    - by user296359
    Hi, I have a question about getting DLL's directory on Windows system. The situation is like this : I have a DLL and an EXE file. The exe file must load the DLL to run. These 2 modules are in different directories. Moreover, the directory of the DLL is changeable. Now I have to get the directory of the DLL in "run time". How could I do this? Thanks in advance.

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