Search Results

Search found 69604 results on 2785 pages for 'file icons'.

Page 34/2785 | < Previous Page | 30 31 32 33 34 35 36 37 38 39 40 41  | Next Page >

  • Overwriting lines in file in C

    - by KáGé
    Hi, I'm doing a project on filesystems on a university operating systems course, my C program should simulate a simple filesystem in a human-readable file, so the file should be based on lines, a line will be a "sector". I've learned, that lines must be of the same length to be overwritten, so I'll pad them with ascii zeroes till the end of the line and leave a certain amount of lines of ascii zeroes that can be filled later. Now I'm making a test program to see if it works like I want it to, but it doesnt. The critical part of my code: file = fopen("irasproba_tesztfajl.txt", "r+"); //it is previously loaded with 10 copies of the line I'll print later in reverse order /* this finds the 3rd line */ int count = 0; //how much have we gone yet? char c; while(count != 2) { if((c = fgetc(file)) == '\n') count++; } fflush(file); fprintf(file, "- . , M N B V C X Y Í U Á É L K J H G F D S A Ú O P O I U Z T R E W Q Ó Ü Ö 9 8 7 6 5 4 3 2 1 0\n"); fflush(file); fclose(file); Now it does nothing, the file stays the same. What could be the problem? Thank you.

    Read the article

  • Delphi Shell IExtractIcon usage and result

    - by Roy M Klever
    What I do: Try to extract thumbnail using IExtractImage if that fail I try to extract icons using IExtractIcon, to get maximum iconsize, but IExtractIcon gives strange results. Problem is I tried to use a methode that extracts icons from an imagelist but if there is no large icon (256x256) it will render the smaller icon at the topleft position of the icon and that does not look good. That is why I am trying to use the IExtractIcon instead. But icons that show up as 256x256 icons in my imagelist extraction methode reports icon sizes as 33 large and 16 small. So how do I check if a large (256x256) icon exists? If you need more info I can provide som sample code. if PThumb.Image = nil then begin OleCheck(ShellFolder.ParseDisplayName(0, nil, StringToOleStr(PThumb.Name), Eaten, PIDL, Atribute)); ShellFolder.GetUIObjectOf(0, 1, PIDL, IExtractIcon, nil, XtractIcon); CoTaskMemFree(PIDL); bool:= False; if Assigned(XtractIcon) then begin GetLocationRes := XtractIcon.GetIconLocation(GIL_FORSHELL, @Buf, sizeof(Buf), IIdx, IFlags); if (GetLocationRes = NOERROR) or (GetLocationRes = E_PENDING) then begin Bmp := TBitmap.Create; try OleCheck(XtractIcon.Extract(@Buf, IIdx, LIcon, SIcon, 32 + (16 shl 16))); Done:= False; Roy M Klever

    Read the article

  • Access violations in strange places when using Windows file dialogs

    - by Robert Oschler
    A long time ago I found out that I was getting access violations in my code due to the use of the Delphi Open File and/or Save File dialogs, which encapsulate the Windows dialogs. I asked some questions on a few forums and I was told that it may have been due to the way some programs add hooks to the shell system that result in DLLs getting injected in every process, some of which can cause havoc with a program. For the record, the programming environment I use is Delphi 6 Professional running on Windows XP 32-bit. At the time I got around it by not using Delphi's Dialog components and instead calling straight into comdlg32.dll. This solved the problem wonderfully. Today I was working with memory mapped files for the first time and sure enough, access violations started cropping up in weird parts of the code. I tried my comdlg32.dll direct calls and this time it didn't help. To isolate the problem as a test I created a list box with the exact same files I was using during testing. These are the exact same test files I was selecting from an Open File dialog and then launching my memory mapped file with. I set things up so that by clicking on a file in the list box, I would use that file in my memory mapped file test instead of calling into a comdlg32.dll dialog function to select a test file. Again, the access violatons vanished. To show you how dramatic a fix it was I went from experiencing an access violation within 1 to 3 trials to none at all. Unfortunately, it's going to bite me later on of course when I do need to use file dialogs. Has anyone else dealt with this issue too and found the real culprit? Did any of you find a solution I could use to fix this problem instead of dancing around it like I am now? Thanks in advance.

    Read the article

  • PHP File Upload using url parameters

    - by Arthur
    Is there a way to upload a file to server using php and the filename in a parameter (instead using a submit form), something like this: myserver/upload.php?file=c:\example.txt Im using a local server, so i dont have problems with filesize limit or upload function, and i have a code to upload file using a form <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "DTD/xhtml1-transitional.dtd"> <html> <body> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" name="fileForm" enctype="multipart/form-data"> File to upload: <table> <tr><td><input name="upfile" type="file"></td></tr> <tr><td><input type="submit" name="submitBtn" value="Upload"></td></tr> </table> </form> <?php if (isset($_POST['submitBtn'])){ // Define the upload location $target_path = "c:\\"; // Create the file name with path $target_path = $target_path . basename( $_FILES['upfile']['name']); // Try to move the file from the temporay directory to the defined. if(move_uploaded_file($_FILES['upfile']['tmp_name'], $target_path)) { echo "The file ". basename( $_FILES['upfile']['name']). " has been uploaded"; } else{ echo "There was an error uploading the file, please try again!"; } } ?> </body> Thanks for the help

    Read the article

  • Most efficient way to write over file after reading

    - by Ryan McClure
    I'm reading in some data from a file, manipulating it, and then overwriting it to the same file. Until now, I've been doing it like so: open (my $inFile, $file) or die "Could not open $file: $!"; $retString .= join ('', <$inFile>); ... close ($inFile); open (my $outFile, $file) or die "Could not open $file: $!"; print $outFile, $retString; close ($inFile); However I realized I can just use the truncate function and open the file for read/write: open (my $inFile, '+<', $file) or die "Could not open $file: $!"; $retString .= join ('', <$inFile>); ... truncate $inFile, 0; print $inFile $retString; close ($inFile); I don't see any examples of this anywhere. It seems to work well, but am I doing it correctly? Is there a better way to do this?

    Read the article

  • Conditionaly strip the last line from a text file

    - by fraXis
    Hello, I posted this yesterday on SO, and I received an answer that works great, but I need to change it around and I don't know how. Here is my original message: I need to strip the last line from a text file. I know how to open and save text files in C#, but how would I strip the last line of the text file? The text file will always be different sizes (some have 80 lines, some have 20). Can someone please show me how to do this? Here is the code that someone gave me to do this (which works fine) //Delete the last line from the file. This line could be 8174, 10000, or anything. This is from SO string tempfile = @"C:\junk_temp.txt"; using (StreamReader reader2 = new StreamReader(newfilename)) { using (StreamWriter writer2 = new StreamWriter(tempfile)) { string line = reader2.ReadLine(); while (!reader2.EndOfStream) { writer2.WriteLine(line); line = reader2.ReadLine(); } // by reading ahead, will not write last line to file } } File.Delete(newfilename); File.Move(tempfile, newfilename); File.Delete(tempfile); How would I change this to only delete the last line of the text file if it is a 4 or 5 digit string (such as 8001 or 99999). If it is anything other than that, such as a %, then I don't want to delete the last line. Can someone please modify the above code to do this for me? Thanks so much.

    Read the article

  • PHP unlink OR rewrite own/current file by itself

    - by Email
    Hi Task: Cut or erase a file after first walk-through. i have an install file called "index.php" which creates another php file. <? /* here some code*/ $fh = fopen($myFile, 'w') or die("can't open file"); $stringData = "<?php \n echo 'hallo, *very very long text*'; \n ?>"; fwrite($fh, $stringData); /*herecut"/ /*here some code */ after the creation of the new file this file is called and i intent to erase the filecreation call since it is very long and only needed on first install. i therefor add to the above code echo 'hallo, *very very long text*'; \n ***$new= file_get_contents('index.php'); \n $findme = 'habanot'; $pos = strpos($new, $findme); if ($pos === false) { $marker='herecut';\n $new=strstr($new,$marker);\n $new='<?php \n /*habanot*/\n'.$new;\n $fh = fopen('index.php', 'w') or die 'cant open file'); $stringData = $new; fwrite($fh, $stringData); fclose($fh);*** ?>"; fwrite($fh, $stringData);]} Isnt there an easier way or a function to modify the current file or even "self destroy" a file after first call? Regards

    Read the article

  • Unable to write to a text file

    - by chrissygormley
    Hello, I am running some tests and need to write to a file. When I run the test's the open = (file, 'r+') does not write to the file. The test script is below: class GetDetailsIP(TestGet): def runTest(self): self.category = ['PTZ'] try: # This run's and return's a value result = self.client.service.Get(self.category) mylogfile = open("test.txt", "r+") print >>mylogfile, result result = ("".join(mylogfile.readlines()[2])) result = str(result.split(':')[1].lstrip("//").split("/")[0]) mylogfile.close() except suds.WebFault, e: assert False except Exception, e: pass finally: if 'result' in locals(): self.assertEquals(result, self.camera_ip) else: assert False When this test run's, no value has been entered into the text file and a value is returned in the variable result. I havw also tried mylogfile.write(result). If the file does not exist is claim's the file does not exist and doesn't create one. Could this be a permission problem where python is not allowed to create a file? I have made sure that all other read's to this file are closed so I the file should not be locked. Can anyone offer any suggestion why this is happening? Thanks

    Read the article

  • Write to the second line of a PHP file

    - by Woz
    I have a php file that I want to add an include path to on the second line. I need to open the file and inset a line of code on line 2. I have tried a few techniques none of which are working but I think it has something to do with the text I am trying to write and possibly not escaping character correctly as I am not too familiar with file writing. So here is the file I want to write to: $file = $_SERVER['DOCUMENT_ROOT'].'/'.$domaindir.'/test.php'; Here is the piece of text I want to insert into the file: $dbfile = "include('".$_SERVER['DOCUMENT_ROOT']."/".$domaindir."/web_".$dbname.".inc.php');"; Then what I was doing was a string replace but all it did was bump the "session_start();" bit to a newline! Can anyone point me in the direction of a tutorial that might tell me how to insert this into the second line of my php file or indeed if anyone has any ideas? I can say for sure that the path to the PHP file is fully tested so i know its not that the file is not being open or written to. Any ideas would be much appreciated. Thanks in advance.

    Read the article

  • Truncating a file while it's being used (Linux)

    - by Hobo
    I have a process that's writing a lot of data to stdout, which I'm redirecting to a log file. I'd like to limit the size of the file by occasionally copying the current file to a new name and truncating it. My usual techniques of truncating a file, like cp /dev/null file don't work, presumably because the process is using it. Is there some way I can truncate the file? Or delete it and somehow associate the process' stdout with a new file? FWIW, it's a third party product that I can't modify to change its logging model. EDIT redirecting over the file seems to have the same issue as the copy above - the file returns to its previous size next time it's written to: ls -l sample.log ; echo > sample.log ; ls -l sample.log ; sleep 10 ; ls -l sample.log -rw-rw-r-- 1 user group 1291999 Jun 11 2009 sample.log -rw-rw-r-- 1 user group 1 Jun 11 2009 sample.log -rw-rw-r-- 1 user group 1292311 Jun 11 2009 sample.log

    Read the article

  • Icons for websites with an <img> tag

    - by Paul Tarjan
    What is the best way to link to the icons of websites? Meaning, given a hostname, what src should I put in an <img tag to link to the 16x16 icon? I was just doing http://<hostname>/favicon.ico. It seems some .ico files aren't liked by different browsers. Chrome seems to like them all, but Safari, FF and IE all have problems with various icons. Example page: http://paultarjan.com

    Read the article

  • Disable menu icons in Visual Studio 2008 AddIn

    - by Wolfgang Ziegler
    I have developed an addin for Visual Studio 2008, which extends the main menu with custom menu items. These menu items have custom images and I finally managed to have them displayed correctly using transparency masks. The only problem that still persists is, that the icons look really ugly and unprofessional, when the menu items are disabled. Instead of getting grayed out smoothly, the icons become flat gray chunks.

    Read the article

  • DRUPAL: replace tags with icons

    - by Patrick
    hi, is there any plugin for Drupal, replacing tags with small icons (see the picture below, the icons are actually small circles, and the different colors are automatically generated. I need to replace the tags with the circles, for each node, and the starting view. Furthermore, when the mouse move over the tags, the tag title should appear as pop-up thanks

    Read the article

  • Java File URI error ?

    - by Frank
    I need to get a file object online, and I know the file is located at : http://nmjava.com/Dir_App_IDs/Dir_GlassPaneDemo/GlassPaneDemo_2010_04_06_15_00_SNGRGLJAMX If I paste it into my browser's url, I'll be able to download this file, now I'm trying to get it with Java, my code looks like this : String File_Url="http://nmjava.com/Dir_App_IDs/Dir_GlassPaneDemo/GlassPaneDemo_2010_04_06_15_00_SNGRGLJAMX"; Object myObject=Get_Online_File(new URI(File_Url)); Object Get_Online_File(URI File_Uri) throws IOException { return readObject(new ObjectInputStream(new FileInputStream(new File(File_Uri)))); } public static synchronized Object readObject(ObjectInput in) throws IOException { Object o; ...... return o; } But I got the following error message : java.lang.IllegalArgumentException: URI scheme is not "file" at java.io.File.<init>(File.java:366) Why ? How to fix it ? Frank

    Read the article

  • C# .Net file in use issue

    - by Dan
    I'm having an issue opening files that have recently been closed by the .Net framework. Basically, what happens is the following: -Read in an XML file using DataSet.ReadXml() -Make some changes to the data -Write out the XML file using DataSet.WriteXml() -Copy the XML file to a new location using File.Copy -FTP the file using a custom control This sequence can intermittently fail either after the WriteXML or the File.Copy with a file in use exception. I'm guessing it could be the Windows write cache not flushing right away. Can anyone confirm that this could be causing my issue? Any solutions to suggest? Thanks, Dan

    Read the article

  • How to create a new File in Qt

    - by GG
    Hello friends, I am a Qt beginner and just got stuck with the problem. I am looking for a file SomePath/NewDirectoryA/NewFile.kml ( NewFile.kml will be the only file in NewDirectoryA, having this directory just to maintain semantics in the project ). If SomePath/NewDirectoryA/NewFile.kml exists then I will use it in my code and If it doesn't exist then I have to create it. If this File doesn't exist then this directory also doesn't exist in SomePath. So If only I have to create a file I can use QFile and open it in ReadWrite or WriteOnly mode. But the problem is I have to create the file with the directory itself. I tried with QFile with file name SomePath/NewDirectoryA/NewFile.kml but it didn't worked. Please suggest me a way in which I can create a new file( NewFile.kml ) in a new directory( NewDirectorA ) at a given location( SomePath ).

    Read the article

  • Bash: Check if file was modified since used in script

    - by Thomas Münz
    I need to check in a script if a file was modified since I read it (another application can modify it in between). According to bash manual there is a "-N" test which should report if a file was modified since last read. I tried it in a small script but it seems like it doesn't work. #!/bin/bash file="test.txt" echo "test" > $file cat $file; if [ -N $file ]; then echo "modified since read"; else echo "not modified since read"; fi I also tried an alternative way by touching another file and using if [ "file1" -nt "file2 ]; but this works only on a seconds accuracy which may under rare conditions not be sufficient. Is there any other bash-inbuilt solution for this problem or I do really need to use diff or md5sum?

    Read the article

  • Perl, deleting an .xml file created and open with IO::File and XML::Writer?

    - by Sho Minamimoto
    So I'm running through a list of things and have code that creates an .xml file with IO::File called $doc, then I make a new writer with XML::Writer(OUTPUT = $doc). More code runs and I build a big xml file with XML::Writer. Then, near the end of the file, I find out if I need this file at all. If I do need it, I just $writer-end(); $doc-close(); but if I don't need it, what should I enter to just delete all data I've stored/saved and move onto the next file? I tried unlink($docpath) (before and after $doc-close()), the file was not deleted.

    Read the article

  • How to concatenate multiple lines of log file into single variable in batch file?

    - by psych
    I have a log file containing a stack trace split over a number of lines. I need to read this file into a batch file and remove all of the lines breaks. As a first step, I tried this: if exist "%log_dir%\Log.log" ( for /F "tokens=*" %%a in ("%log_dir%\Log.log") do @echo %%a ) My expectation was that this would echo out each line of the log file. I was then planning to concatenate these lines together and set that value in a variable. However, this code doesn't do what I would expect. I have tried changing the value of the options for delims and tokens, but the only output I can get is the absolute path to the log file and nothing from the contents of this file. How can I set a variable to be equal to the lines of text in a file with the line breaks removed?

    Read the article

  • How to know the file type of the file which you are forcing the user to download?

    - by Starx
    I am trying to force the user to download a file. For that my script is: $file = "file\this.zip"; header("Cache-Control: public"); header("Content-Description: File Transfer"); header("Content-Disposition: attachment; filename=$file"); header("Content-Type: application/zip"); //This is what I need header("Content-Transfer-Encoding: binary"); readfile($file); The files I am going to upload my not be .zip all the time so I want to know the content type of the image I am going to receive in $file. How to accomplish this

    Read the article

  • How can I insert a line at the beginning of a file with Perl's Tie::File?

    - by thebourneid
    I'm trying to insert/add a line 'COMMENT DUMMY' at the beginnig of a file as a first row if /PATTERN/ not found. I know how to do this with OPEN CLOSE function. Probably after reading the file it should look something like this: open F, ">", $fn or die "could not open file: $!"; ; print F "COMMENT DUMMY\n", @array; close F; But I have a need to implement this with the use of the Tie::File function and don't know how. use strict; use warnings; use Tie::File; my $fn = 'test.txt'; tie my @lines, 'Tie::File', $fn or die "could not tie file: $!"; untie @lines;

    Read the article

  • Create Zip File In Windows and Extract Zip File In Linux

    - by Yan Cheng CHEOK
    I had created a zip file (together with directory) under Windows as follow : package sandbox; import java.io.File; import java.io.FileInputStream; import java.io.FileOutputStream; import java.io.IOException; import java.util.zip.ZipEntry; import java.util.zip.ZipOutputStream; /** * * @author yan-cheng.cheok */ public class Main { /** * @param args the command line arguments */ public static void main(String[] args) { // These are the files to include in the ZIP file String[] filenames = new String[]{"MyDirectory" + File.separator + "MyFile.txt"}; // Create a buffer for reading the files byte[] buf = new byte[1024]; try { // Create the ZIP file String outFilename = "outfile.zip"; ZipOutputStream out = new ZipOutputStream(new FileOutputStream(outFilename)); // Compress the files for (int i=0; i<filenames.length; i++) { FileInputStream in = new FileInputStream(filenames[i]); // Add ZIP entry to output stream. out.putNextEntry(new ZipEntry(filenames[i])); // 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(); } catch (IOException e) { e.printStackTrace(); } } } The newly created zip file can be extracted without problem under Windows, by using http://www.exampledepot.com/egs/java.util.zip/GetZip.html However, I realize if I extract the newly created zip file under Linux, using http://www.exampledepot.com/egs/java.util.zip/GetZip.html, I will get a file named "MyDirectory\MyFile.txt" instead of MyFile.txt being placed under folder MyDirectory. I try to solve the problem by changing the zip file creation code to String[] filenames = new String[]{"MyDirectory" + "/" + "MyFile.txt"}; But, is this an eligible solution, by hard-coded the seperator? Will it work under Mac OS? (I do not have a Mac to try out)

    Read the article

< Previous Page | 30 31 32 33 34 35 36 37 38 39 40 41  | Next Page >