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  • List all files from a directory recursively with Java

    - by Hultner
    Okay I got this function who prints the name of all files in a directory recursively problem is that it's very slow and it gets the stuff from a network device and with my current code it has to access the device time after time. What I would want is to first load all the files from the directory recursively and then after that go through all files with the regex to filter out all the files I don't want. Unless anyone got a better suggestion. I've never before done anything like this. public static printFnames(String sDir){  File[] faFiles = new File(sDir).listFiles();  for(File file: faFiles){ if(file.getName().matches("^(.*?)")){   System.out.println(file.getAbsolutePath()); }   if(file.isDirectory()){     printFnames(file.getAbsolutePath());   }  } } This is just a test later on I'm not going to use the code like this, instead I'm going to add the path and modification date of every file which matches an advanced regex to an array.

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  • Optimal directory structure for filesystem

    - by Pankaj
    We have large scale web application which has millions of customer. Each customer can have document based on document type. We may have 20-30 types of documents. We are planning to use GlusterFS for storing these documents. I'm trying to find out what are the limitations of Gluster as far as number of files/directories ? Do we need to have hierarchical directory structure ? What would be the optimal directory structure ? Does this make sense - CustmerId Documenttype File1 File2

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  • phpunit - generate tests for all the classes from a directory

    - by joe
    Hi! I have a directoy structure, and all the classes of the business logic are placed in the app_dir/lib/ directory. I would like to generate unit tests for all the classes from this lib/ folder. The problem is, that I haven't found any option to specify the source directory, only the source file: from app_dir: phpunit --skeleton-class lib/ Error: "lib/.php" could not be opened. Is it the only solution to write my own php script, which iterates through the /lib folder and calls the skeleton generator for every file found? And how can I specify the output folder, where all the generated test files are placed? Thanx, Joe

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  • How do I make a symlink to every directory in the current directory that has the same name but has u

    - by Jason Baker
    For instance, I suppose I have a directory that contains the following folders foo_bar baz What I would like to have is a bash command that will make a symlink foo-bar to foo_bar so it would look like this: foo-bar foo_bar baz I'm pretty sure I can write a Python script to do this, but I'm curious if there's a way to do this with bash. Here's where I'm stuck: ls -1 | grep _ | xargs -I {} ln -s {} `{} | sed 's/_/-/'` What I'm trying to do is run the command ln -s with the first argument being the directory name and the second argument being that name passed through sed s/_/-/. Is there another way to do this?

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  • apache on windows virtual directory config help

    - by sprugman
    I'm running Apache on Windows XP via Xampplite, and could use help configuring my virtual directory. Here's what I'm hoping to do on my dev box: I want my source files to live outside of the xampp htdocs dir on my local machine I can access the project at http://myproject others on my local network can access the project at my.ip.address/myproject keep localhost pointing to the xampp's htdocs folder so I can easily add other projects. I've got 1 & 2 working by editing the windows hosts file, and adding a virtual directory in xampp's apache\conf\extra\httpd-vhosts.conf file. I don't immediately see how to do 3 without messing up 4.

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  • PHP readdir(): 3 is not a valid Directory resource

    - by Jordan
    <?php function convert(){ //enable error reporting for debugging error_reporting(E_ALL | E_STRICT); //convert pdf's to html using payroll.sh script and //pdftohtml command line tool $program = "payroll.sh"; $toexec="sh /var/www/html/tmp/" . $program . " 2>&1"; $exec=shell_exec($toexec); //display message from payroll.sh //echo $exec; //echo ('<br/>'); } function process(){ $dir = '/var/www/html/tmp/converted'; //echo ('one'); if (is_dir($dir)) { //echo ('two'); if ($dh = opendir($dir)) { //echo ('three'); while (($file = readdir($dh)) !== false) { //echo ('four'); if ($file != "." && $file != ".."){ echo 'opening file: '; echo $file; echo ("<br/>"); $fp = fopen('/var/www/html/tmp/converted/' . $file, 'r+'); $count = 0; //while file is not at the EOF marker while (!feof($fp)) { $line = fgets($fp); if($count==21) { $employeeID = substr($line,71,4); echo 'employee ID: '; echo $employeeID; echo ('<br/>'); //echo ('six'); $count++; } else if($count==30) { $employeeDate = substr($line,71,10); echo 'employee Date: '; echo $employeeDate; echo ('<br/>'); //echo ('seven'); $count++; } else { //echo ('eight'); //echo ('<br/>'); $count++; } } fclose($fp); closedir($dh); } } } } } convert(); process(); ?> I am setting up a php script that will take a paystub in pdf format, convert it to html, then import it into Drupal after getting the date and employee ID. The code only seems to process the first file in the directory then it gives me this: opening file: dd00000112_28_2010142011-1.html employee ID: 9871 employee Date: 12/31/2010 Warning: readdir(): 3 is not a valid Directory resource in /var/www/html/pay.mistequaygroup.com/payroll.php on line 29 The '3' in the error really confuses me, and google is not helping much. Could it be the 3rd iteration of the loop? The only files in the directory reddir() is scanning are the .html files waiting to be processed. Any ideas? Also, how does my code look? I'm fairly new to doing any real programming and I don't get too much input around work.

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  • parse this directory path without losing slash

    - by PPTim
    hi, I have a wxPython application. I am taking in a directory path from a textbox using GetValue(). I notice that while trying to parse in the directory path "C:\Documents and Settings\UserName\Desktop\InputFile.xls", python sees the string as 'C:\\Documents and Settings\UserName\\Desktop\\InputFile.xls' (missing a slash between "Settings" and "UserName). Why is it that only that slash is not correctly escaped? Once the string has been changed to 'C:\Documents and Settings\UserName\Desktop\InputFile.xls', is there a type conversion or function that can does this properly? Thanks.

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  • BASH Script to cd to directory with spaces in pathname

    - by Rails Newbie
    Argggg. I've been struggling with this stupid problem for days and I can't find an answer. I'm using BASH on Mac OS X and I'd like to create a simple executable script file that would change to another directory when it's run. However, the path to that directory has spaces in it. How the heck do you do this? This is what I have... Name of file: cdcode File contents: cd ~/My Code Now granted, this isn't a long pathname, but my actual pathname is five directories deep and four of those directories have spaces in the path. BTW, I've tried cd "~/My Code" and cd "~/My\ Code" and neither of these worked. If you can help, THANKS! This is driving me crazy!!

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  • Create Directory, 'cd' to it and download a file pipeline in Perl

    - by neversaint
    I have a file that looks like this: ftp://url1/files1.tar.gz dir1 ftp://url2/files2.txt dir2 .... many more... What I want to do are these steps: Create directory based on column 2 Unix 'cd' to that directory Download file with 'wget' based on column1 But how come this approach of mine doesn't work while(<>) { chomp; my ($url,$dir) = split(/\t/,$_); system("mkdir $dir"); system("cd $dir"); # Fail here system("wget $url"); # here too } What's the right way to do it?

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  • Eclipse buildpath automatically taking all JARs of a internal directory

    - by Niko
    How do I configure my project buildpath to have a set of .jar files located in the same directory automatically included in the buildpath ? Meaning that adding a new .jar file to this directory (and refreshing the project) updates the buildpath ? Rem : I am not working in a Webapp but in a standalone Java app. I know that it is possible in a Dynamic Web Project to have all the .jars located in WEB-INF/lib to be included in the build path. Is it possible to do kind of the same include but in standalone app ? I am using Eclipse 3.4

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  • Moving directory after compilation of R

    - by CravingSpirit
    I compiled R in /tmp/R-3.0.0 and then moved it to /home/user/opt/R-3.0.0, then I got an error when executing R: /home/kaiyin/opt/R-3.0.0/bin/R: line 236: /tmp/R-3.0.0/etc/ldpaths: No such file or directory ERROR: R_HOME ('/tmp/R-3.0.0') not found If I export R_HOME='/home/kaiyin/opt/R-3.0.0', it still gives almost the same error: WARNING: ignoring environment value of R_HOME /home/kaiyin/opt/R-3.0.0/bin/R: line 236: /tmp/R-3.0.0/etc/ldpaths: No such file or directory ERROR: R_HOME ('/tmp/R-3.0.0') not found Is there a way to solve this, or do I have to recompile it?

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  • Java import from other directory

    - by heldopslippers
    Hi People! I am building a Enterprise Service Bus (ESB) with Java. I won't get into details But I have to build multiple servers who make use of the same classes. I have the following directory structure: /server1 -Main.java /server2 -Main.java /com -Database.java I want to import from the Main.java class for example the Database.class. But of course the following statements won't work: import com.Database; I am working with the javac compiler in the command line (so not eclipse stuff or whatever. just TextMate and the command line). And I found a (pretty stupid) solution by creating a symbolic link in the servers to the com directory. But that is not really an ideal solution. Does anybody have a better one?? THANXS in advanced!! :D

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  • Viewing directory containing MIME encoded email messages

    - by Mark
    I have an application which generates and sends MIME encoded messages (javax.mail.internet.MimeMessage) through an SMTP server. As part of the development process only, I'd like to be able to view these messages rather than send them (I know the sending works just fine, but there are restrictions on the domains within the dev environment which makes it a little difficult) I thought the easiest way would be to save the text for each message to a directory, then point "an app" at the directory and check them over. So the question is, what would be a good app to use? Is it as simple as configuring Outlook or another email client to do it? Thanks

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  • git - is there a way to get only required files in the working directory

    - by spoonboy
    I'm new to git and trying to use it with a project that has many (several hundreds) sources. The problem I have is that git is extracting all the project's sources to my working directory when doing checkout. This makes a lot of mess as I have to jump between the files and can unintentionally change/corrupt files that I wasn't even planning to change. I would prefer to extract only sources that I'm going to modify and then work with them. So, is there a way to tell git that I only going to work with specific sources, and so, that only these sources would be extracted to the working directory? Note, that this is not a partial checkout or something like this. I'm ok to checkout the whole branch. It's more about organising a working folder. Thanks.

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  • Error becuase Virtual directory not being configured as an application in IIS

    - by Cipher
    Hi, I was trying to install a CMS in a folder in my website. After the installation on trying to run, it shows this error: Error 14 It is an error to use a section registered as allowDefinition='MachineToApplication' beyond application level. This error can be caused by a virtual directory not being configured as an application in IIS. E:\Users\Sarin\Documents\Visual Studio 2010\WebSites\WebAssist\blog\web.config 61 I added the webiste as a Virtual Directory and also converted that to application. On trying to browse this application, the following error occurs as shown in the screenshot: http://i.imgur.com/jcRJe.jpg How to make this work?

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  • Write file in sub-directory in Android

    - by Davide Vosti
    I'm trying to save a file in a subdirectory in Android 1.5. I can successfully create a directory using _context.GetFileStreamPath("foo").mkdir(); (_context is the Activity where I start the execution of saving the file) but then if I try to create a file in foo/ by _context.GetFileStreamPath("foo/bar.txt"); I get a exception saying I can't have directory separator in a file name ("/"). I'm missing something of working with files in Android... I thought I could use the standard Java classes but they don't seem to work... I searched the Android documentation but I couldn't fine example and google is not helping me too... I'm asking the wrong question (to google)... Can you help me out with this? Thank you!

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