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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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  • Save PHP variables to a text file

    - by Ajith
    I was wondering how to save PHP variables to a txt file and then retrieve them again. Example: There is an input box, after submitted the stuff that was written in the input box will be saved to a text file. Later on the results need to be brought back as a variable. So lets say the variable is $text I need that to be saved to a text file and be able to retrieve it back again. Hope it makes sense, Thanks in advance!!!

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  • Reading parameters from External file - C#

    - by mouthpiec
    I am writing an application using C# and I would like to read some parameters from an external file like for example a text file. The parameters will be saved in the file in the form of parA = 5 parB = hello etc Can you pleas suggest a way how I can do this?

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  • php,codigniter, upload the zip file problem

    - by user345804
    function upload( &$data = array() ) { $config['upload_path'] = 'system/application/orginalimage/'; $config['allowed_types'] = 'gif|jpg|png|zip'; $config['allowed_type'] = 'application/zip'; $config['allowed_type'] = 'application/x-zip-compressed'; $config['allowed_type'] = 'application/x-compress'; $config['allowed_type'] = 'application/x-compressed'; $config['allowed_type'] = 'application/octet-stream'; $config['allowed_type'] = 'multipart/x-zip'; $config['max_size'] = '100'; $config['max_width'] = '1024'; $config['max_height'] = '768'; $this->load->library('upload', $config); if ( $this->upload->do_upload( 'uploadimage')) { $data = $this->upload->data() ; return true ; } return false ; } uploading a zip file is not working help me

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  • Tunnel Failed at the time of Upload file to FTP

    - by Karthick
    File upload is works fine from my simulator (blackberry 8830).It upload the file to FTP Server. But in the device when I try to upload file to FTP server it gives the alert “Tunnel Failed “. I am using StreamConnection sc = (StreamConnection) Connector.open(url); How to solve this issue. Can anyone help me???

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  • Reading line by line from a file in C

    - by mh234
    What I am trying to do is print out the contents of a file line by line. I run the program in terminal by doing: ./test testText.txt. When I do this, random characters are printed out but not what is in the file. The text file is located in the same folder as the makefile. What's wrong? #include <stdio.h> FILE *fp; int main(int argc, char *argv[]) { char line[15]; fp = fopen(*argv, "r"); while((fgets(line, 15, fp)) != NULL) { printf(line); printf("\n"); } }

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  • [PHP] What exactly does it mean when $_FILES is empty?

    - by Mike
    I am working on a PHP upload script and when testing my error checks, I attempted to upload a 17MB TIFF file. When I do this the $_FILES array is empty. The script works fine for what I need it to do, which is to upload JPEG files. My solution is to test if $_FILES is empty or not before continuing with the upload script. Can anybody explain why $_FILES is empty when a TIFF is attempted to be uploaded? Is my solution, to check if $_FILES is empty or not, an okay one? Does this have something to do with settings in php.ini?

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  • jquery-autocomplete does not work with my django app.

    - by HWM-Rocker
    Hi everybody, I have a problem with the jquery-autocomplete pluging and my django script. I want an easy to use autocomplete plugin. And for what I see this (http://code.google.com/p/jquery-autocomplete/) one seems very usefull and easy. For the django part I use this (http://code.google.com/p/django-ajax-selects/) I modified it a little, because the out put looked a little bit weired to me. It had 2 '\n' for each new line, and there was no Content-Length Header in the response. First I thought this could be the problem, because all the online examples I found had them. But that was not the problem. I have a very small test.html with the following body: <body> <form action="" method="post"> <p><label for="id_tag_list">Tag list:</label> <input id="id_tag_list" name="tag_list" maxlength="200" type="text" /> </p> <input type="submit" value="Submit" /> </form> </body> And this is the JQuery call to add autocomplete to the input. function formatItem_tag_list(bla,row) { return row[2] } function formatResult_tag_list(bla,row) { return row[1] } $(document).ready(function(){ $("input[id='id_tag_list']").autocomplete({ url:'http://gladis.org/ajax/tag', formatItem: formatItem_tag_list, formatResult: formatResult_tag_list, dataType:'text' }); }); When I'm typing something inside the Textfield Firefox (firebug) and Chromium-browser indicates that ther is an ajax call but with no response. If I just copy the line into my browser, I can see the the response. (this issue is solved, it was a safety feature from ajax not to get data from another domain) For example when I am typing Bi in the textfield, the url "http://gladis.org/ajax/tag?q=Bi&max... is generated. When you enter this in your browser you get this response: 4|Bier|Bier 43|Kolumbien|Kolumbien 33|Namibia|Namibia Now my ajax call get the correct response, but there is still no list showing up with all the possible entries. I tried also to format the output, but this doesn't work either. I set brakepoints to the function and realized that they won't be called at all. Here is a link to my minimum HTML file http://gladis.org/media/input.html Has anybody an idea what i did wrong. I also uploaded all the files as a small zip at http://gladis.org/media/example.zip. Thank you for your help! [Edit] here is the urls conf: (r'^ajax/(?P<channel>[a-z]+)$', 'ajax_select.views.ajax_lookup'), and the ajax lookup channel configuration AJAX_LOOKUP_CHANNELS = { # the simplest case, pass a DICT with the model and field to search against : 'tag' : dict(model='htags.Tag', search_field='text'), } and the view: def ajax_lookup(request,channel): """ this view supplies results for both foreign keys and many to many fields """ # it should come in as GET unless global $.ajaxSetup({type:"POST"}) has been set # in which case we'll support POST if request.method == "GET": # we could also insist on an ajax request if 'q' not in request.GET: return HttpResponse('') query = request.GET['q'] else: if 'q' not in request.POST: return HttpResponse('') # suspicious query = request.POST['q'] lookup_channel = get_lookup(channel) if query: instances = lookup_channel.get_query(query,request) else: instances = [] results = [] for item in instances: results.append(u"%s|%s|%s" % (item.pk,lookup_channel.format_item(item),lookup_channel.format_result(item))) ret_string = "\n".join(results) resp = HttpResponse(ret_string,mimetype="text/html") resp['Content-Length'] = len(ret_string) return resp

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  • SFTP jail & Keeping file ownership the same / File owner per folder

    - by Dragonshadow
    I want to setup a jailed SFTP account for a subfolder of another user's home folder, but want the owner of everything in that subfolder to stay the same, including new files and folders uploaded and created by the sftp user, while still allowing access to the files and folders of that subfolder as if the SFTP user was the parent user. rawny bawb-sftp /home/rawny <- rawny owns this /home/rawny/sftp <- rawny owns this too, but bawb-sftp can upload to it, edit files, etc bawb-sftp uploads a file /home/rawny/sftp/lol.txt rawny should still own the file, as if he made it in the first place, even though bawb-sftp was the one that uploaded it. Basically I guess I'm asking for an sftp jail that acts as a highly limited passthrough/puppet for another user?

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  • DIR $file "File Not Found" vs DIR $filedir shows it....not permissions, not USB

    - by Kev
    I was having this problem before on a USB drive, but now it's happening on my main RAID5-backed hard disk: 2013-10-17 9:37 C:\>dir "C:\Shares\Shared\Reference\Safety Management System\Vid eo CD\AutoPlay\Docs\Manuel*" Volume in drive C has no label. Volume Serial Number is 3C18-E114 Directory of C:\Shares\Shared\Reference\Safety Management System\Video CD\AutoP lay\Docs 2003-09-09 11:29 PM 1,056,768 Manuel d'intervention d'urgence MFC.doc 2004-06-20 10:36 PM 139,849 Manuel d'intervention d'urgence MFC.pdf 2 File(s) 1,196,617 bytes 0 Dir(s) 196,068,691,968 bytes free 2013-10-17 9:38 C:\>dir "C:\Shares\Shared\Reference\Safety Management System\Vid eo CD\AutoPlay\Docs\Manuel d'intervention d'urgence MFC.doc" Volume in drive C has no label. Volume Serial Number is 3C18-E114 Directory of C:\Shares\Shared\Reference\Safety Management System\Video CD\AutoP lay\Docs File Not Found 2013-10-17 9:38 C:\> This is from a Command Prompt window where I went to Properties and told it I wanted to modify who it ran as. I opened it, had it run as me with the "restricted access" unchecked, then ran the above. The file in question has the following ACLs: Administrators, SYSTEM, and OurCompanyUsers. All three have full control of everything. Nobody has any Deny bits set. I am a member of Administrators. So I don't believe it's a permissions issue. It's not a USB drive, so this time there is no question of USB hardware. Windows Server 2003 Standard Edition SP2. What does this mean? Is this more likely a hardware or software problem?

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  • Is there such a thing as a file hosted container which deduplicates data held within?

    - by Mallow
    Background I have backups of a website which stores all of it's data into a single file. This file is several gigs large and I have many different backups of this file. Most of the data within is mostly the same plus whatever was added or changed to it. I want to keep all the concurrent backups I've made through the years in case I find a horrible surprise of data corruption along the line. However storing a 10gig file every month gets expensive. Seeking Solution I've often thought about different ways of alleviating this problem. One thought that comes up very often combines the idea of a duplicating file system which doesn't require it's own partitioned volume on a hard drive. Something like what truecrypt does, what it calls, "file hosted containers" which when using the truecrypt program allows you to mount and dismount that volume as a regular hard drive. Question Is there a virtual hard drive mounter which uses file-based container which uses data deduplicaiton file system? (This question is a little awkward to put into words, if you have a better idea on how to ask this question please feel free to help out.)

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  • How should I configure my Apache Hosts File to serve a different site for localhost than for my domain/publicip?

    - by rofls
    I'm trying to test out a LAMP (with PHP5 specifically) setup with Django already serving a website. I want to do the PHP stuff on localhost for now, so that when I do something like this: curl http://localhost/database/script.php?var=1, I get a response from the php server. Right now I'm getting a Django error. I tried something like this in the default file in sites-available: Listen 80 <VirtualHost aaa.bbb.ccc.ddd> ServerName localhost DocumentRoot /home/phpsite </VirtualHost> where aaa.bbb.ccc.ddd is the local ip address, and changing my actual site's settings to specify the public ip, like this: Listen 80 <VirtualHost www.xxx.yyy.zzz> ServerName mysite.com DocumentRoot /srv/www/mysite WSGIScriptAlias / /srv/www/mysite.wsgi </VirtualHost> but then I start getting all kinds of errors when I start apache, such as port ::[80] is already in use or something. I noticed that the hosts file that's located in /etc/apache2/ is apparently pointing everything to mysite.com, including my local ip as well as 127.0.0.1 and 127.0.1.1; Do I need to change the configuration there too?

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  • Java compiler error: "cannot find symbol" when trying to access local variable

    - by HH
    $ javac GetAllDirs.java GetAllDirs.java:16: cannot find symbol symbol : variable checkFile location: class GetAllDirs System.out.println(checkFile.getName()); ^ 1 error $ cat GetAllDirs.java import java.util.*; import java.io.*; public class GetAllDirs { public void getAllDirs(File file) { if(file.isDirectory()){ System.out.println(file.getName()); File checkFile = new File(file.getCanonicalPath()); }else if(file.isFile()){ System.out.println(file.getName()); File checkFile = new File(file.getParent()); }else{ // checkFile should get Initialized at least HERE! File checkFile = file; } System.out.println(file.getName()); // WHY ERROR HERE: checkfile not found System.out.println(checkFile.getName()); } public static void main(String[] args) { GetAllDirs dirs = new GetAllDirs(); File current = new File("."); dirs.getAllDirs(current); } }

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  • Visual Studio 2008 project file does not load because of an unexpected encoding change.

    - by Xenan
    In our team we have a database project in visual Studio 2008 which is under source control by Team Foundation Server. Every two weeks or so, after one co-worker checks in, the project file won't load on the other developers machines. The error message is: The project file could not be loaded. Data at the root level is invalid. Line 1, position 1. When I look at the project file in Notepad++, the file looks like this: ??<NUL?NULxNULmNULlNUL NULvNULeNULrNULsNULiNULoNULnNUL ... and so on (you can see <?xml version in this) whereas an normal project file looks like: <?xml version="1.0" encoding="utf-16"?> ... So probably something is wrong with the encoding of the file. This is a problem for us because it turns out to be impossible to get the file encoding correct again. The 'solution' is to throw away the project file an get the last know working version from source control. According to the file, the encoding should be UTF-16. According to Notepad++, the corrupted file is actually UTF-8. My questions are: Why is Visual Studio messing up the encoding of the project file, apparently at random times and at random machines? What should we do to prevent this? When it has happened, is there a possibility to restore the current file in the correct encoding instead of pulling an older version from source control? As a last note: the problem is with one single project file, all other project files don't expose this problem.

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  • File.Move does not inherit permissions from target directory?

    - by Joseph Kingry
    In case something goes wrong in creating a file, I've been writing to a temporary file and then moving to the destination. Something like: var destination = @"C:\foo\bar.txt"; var tempFile = Path.GetTempFileName(); using (var stream = File.OpenWrite(tempFile)) { // write to file here here } string backupFile = null; try { var dir = Path.GetDirectoryName(destination); if (!Directory.Exists(dir)) { Directory.CreateDirectory(dir); Util.SetPermissions(dir); } if (File.Exists(destination)) { backupFile = Path.Combine(Path.GetTempPath(), new Guid().ToString()); File.Move(destination, backupFile); } File.Move(tempFile, destination); if (backupFile != null) { File.Delete(backupFile); } } catch(IOException) { if(backupFile != null && !File.Exists(destination) && File.Exists(backupFile)) { File.Move(backupFile, destination); } } The problem is that the new "bar.txt" in this case does not inherit permissions from the "C:\foo" directory. Yet if I create a file via explorer/notepad etc directly in the "C:\foo" there's no issues, so I believe the permissions are correctly set on "C:\foo". Update Found Inherited permissions are not automatically updated when you move folders, maybe it applies to folders as well. Now looking for a way to force an update of file permissions. Is there a better way overall of doing this?

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  • Can't access my files in ASP.NET web site

    - by jumbojs
    I'm having a very difficult time. I am running windows 2008 server, I have an Able Commerce site using ASP.NET with C#. I'm writing an automated task that will ftp some xml files down into a local directory on our web server and then the program parses the xml file and saves information to our database. The problem, once I save the files to our local directory, my program has no access to the files. The NETWORK SERVICE user permissions isn't being inherited by the xml files so my program can't do anything with them. I can manually change the permissions, but this wouldn't be automated and won't work. How can I get this to work? help please, it's very frustrating.

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  • XML File as Excel file.

    - by FrustratedWithFormsDesigner
    I have a number of reports that I run against my database that need to eventually go to the end-users as Excel spreadsheets. Initially, I was creating text reports, but the steps to convert the text to a spreadsheet were a bit cumbersome. There were too many steps to import text to the spreadsheet, and multi-line text rows were imported as individual rows in Excel (which was incorrect). Currently, I am generating simple XML saving the file with an ".xls" extension. This works better, but there is still the problem of Excel prompting the user with an XML import dialogue every time they open the file, and then having to save a new file if they add notes or change the layout to the file (which they almost certainly will be doing). Sample "xls" file: <?xml version="1.0" standalone="yes"?> <report_rows> <row> <NAME>Test Data</NAME> <COUNT>345</COUNT> </row> <!-- many more row elements... --> </report_rows> Is there any way to add markup to the file to hint to Excel how it should import and handle the file? Ideally, the end user should be able to open and save the file like any othe spreadsheet they create directly from Excel. Is this even possible? UPDATE: We are running Office 2003 here. UPDATE: The XML is generated from a sqlplus script, no option to use C#/.NET here.

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  • General question: Filesystem or database?

    - by poeschlorn
    Hey guys, i want to create a small document management system. there are several users who store their files. each file which is uploaded contains an info which user uploaded it and the document content itself. In a view there are displayed all files of ONE specific user, ordered by date. What would be better: 1) giving the documents a name or metadata(XML) which contain the date and user (and iterate through them to get the metadata) or 2) giving the files a random/unique name and store metadata in a DB? something like this: date | user | filename What would you say and why? The used programming language is java and the DB is MySQL.

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  • What would happen if a same file being read and appended at the same time(python programming)?

    - by Shane
    I'm writing a script using two separate thread one doing file reading operation and the other doing appending, both threads run fairly frequently. My question is, if one thread happens to read the file while the other is just in the middle of appending strings such as "This is a test" into this file, what would happen? I know if you are appending a smaller-than-buffer string, no matter how frequently you read the file in other threads, there would never be incomplete line such as "This i" appearing in your read file, I mean the os would either do: append "This is a test" - read info from the file; or: read info from the file - append "This is a test" to the file; and such would never happen: append "This i" - read info from the file - append "s a test". But if "This is a test" is big enough(assuming it's a bigger-than-buffer string), the os can't do appending job in one operation, so the appending job would be divided into two: first append "This i" to the file, then append "s a test", so in this kind of situation if I happen to read the file in the middle of the whole appending operation, would I get such result: append "This i" - read info from the file - append "s a test", which means I might read a file that includes an incomplete string?

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  • How to Upload a file from client to server using OFBIZ?

    - by SIVAKUMAR.J
    I'm new to ofbiz so try to keep your answer as simple as possibly. If you can give examples that would be kind. My problem is I created a project inside the ofbiz/hot-deploy folder namely productionmgntSystem. Inside the folder ofbiz\hot-deploy\productionmgntSystem\webapp\productionmgntSystem I created a file app_details_1.ftl. The following are the code of this file <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1"> <title>Insert title here</title> <script TYPE="TEXT/JAVASCRIPT" language=""JAVASCRIPT"> function uploadFile() { //alert("Before calling upload.jsp"); window.location='<@ofbizUrl>testing_service1</@ofbizUrl>' } </script> </head> <!-- <form action="<@ofbizUrl>testing_service1</@ofbizUrl>" enctype="multipart/form-data" name="app_details_frm"> --> <form action="<@ofbizUrl>logout1</@ofbizUrl>" enctype="multipart/form-data" name="app_details_frm"> <center style="height: 299px; "> <table border="0" style="height: 177px; width: 788px"> <tr style="height: 115px; "> <td style="width: 103px; "> <td style="width: 413px; "><h1>APPLICATION DETAILS</h1> <td style="width: 55px; "> </tr> <tr> <td style="width: 125px; ">Application name : </td> <td> <input name="app_name_txt" id="txt_1" value=" " /> </td> </tr> <tr> <td style="width: 125px; ">Excell sheet &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;: </td> <td> <input type="file" name="filename"/> </td> </tr> <tr> <td> <!-- <input type="button" name="logout1_cmd" value="Logout" onclick="logout1()"/> --> <input type="submit" name="logout_cmd" value="logout"/> </td> <td> <!-- <input type="submit" name="upload_cmd" value="Submit" /> --> <input type="button" name="upload1_cmd" value="Upload" onclick="uploadFile()"/> </td> </tr> </table> </center> </form> </html> the following coding is present in the file ofbiz\hot-deploy\productionmgntSystem\webapp\productionmgntSystem\WEB-INF\controller.xml ...... ....... ........ <request-map uri="testing_service1"> <security https="true" auth="true"/> <event type="java" path="org.ofbiz.productionmgntSystem.web_app_req.WebServices1" invoke="testingService"/> <response name="ok" type="view" value="ok_view"/> <response name="exception" type="view" value="exception_view"/> </request-map> .......... ............ .......... <view-map name="ok_view" type="ftl" page="ok_view.ftl"/> <view-map name="exception_view" type="ftl" page="exception_view.ftl"/> ................ ............. ............. The following are the coding present in the file ofbiz\hot-deploy\productionmgntSystem\src\org\ofbiz\productionmgntSystem\web_app_req\WebServices1.java package org.ofbiz.productionmgntSystem.web_app_req; import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; import java.io.DataInputStream; import java.io.FileOutputStream; import java.io.IOException; public class WebServices1 { public static String testingService(HttpServletRequest request, HttpServletResponse response) { //int i=0; String result="ok"; System.out.println("\n\n\t*************************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response)- Start"); String contentType=request.getContentType(); System.out.println("\n\n\t*************************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response)- contentType : "+contentType); String str=new String(); // response.setContentType("text/html"); //PrintWriter writer; if ((contentType != null) && (contentType.indexOf("multipart/form-data") >= 0)) { System.out.println("\n\n\t**********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) after if (contentType != null)"); try { // writer=response.getWriter(); System.out.println("\n\n\t**********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) - try Start"); DataInputStream in = new DataInputStream(request.getInputStream()); int formDataLength = request.getContentLength(); byte dataBytes[] = new byte[formDataLength]; int byteRead = 0; int totalBytesRead = 0; //this loop converting the uploaded file into byte code while (totalBytesRead < formDataLength) { byteRead = in.read(dataBytes, totalBytesRead,formDataLength); totalBytesRead += byteRead; } String file = new String(dataBytes); //for saving the file name String saveFile = file.substring(file.indexOf("filename=\"") + 10); saveFile = saveFile.substring(0, saveFile.indexOf("\n")); saveFile = saveFile.substring(saveFile.lastIndexOf("\\")+ 1,saveFile.indexOf("\"")); int lastIndex = contentType.lastIndexOf("="); String boundary = contentType.substring(lastIndex + 1,contentType.length()); int pos; //extracting the index of file pos = file.indexOf("filename=\""); pos = file.indexOf("\n", pos) + 1; pos = file.indexOf("\n", pos) + 1; pos = file.indexOf("\n", pos) + 1; int boundaryLocation = file.indexOf(boundary, pos) - 4; int startPos = ((file.substring(0, pos)).getBytes()).length; int endPos = ((file.substring(0, boundaryLocation)).getBytes()).length; //creating a new file with the same name and writing the content in new file FileOutputStream fileOut = new FileOutputStream("/"+saveFile); fileOut.write(dataBytes, startPos, (endPos - startPos)); fileOut.flush(); fileOut.close(); System.out.println("\n\n\t**********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) - try End"); } catch(IOException ioe) { System.out.println("\n\n\t*********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) - Catch IOException"); //ioe.printStackTrace(); return("exception"); } catch(Exception ex) { System.out.println("\n\n\t*********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) - Catch Exception"); return("exception"); } } else { System.out.println("\n\n\t********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) else part"); result="exception"; } System.out.println("\n\n\t*************************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response)- End"); return(result); } } I want to upload a file to the server. The file is get from user " tag in the "app_details_1.ftl" file & it is updated into the server by using the method "testingService(HttpServletRequest request, HttpServletResponse response)" in the class "WebServices1". But the file is not uploaded. Give me a good solution for uploading a file to the server.

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  • Activate a python virtual environment using activate_this.py in a fabfile on Windows

    - by Rudy Lattae
    I have a Fabric task that needs to access the settings of my Django project. On Windows, I'm unable to install Fabric into the project's virtualenv (issues with Paramiko + pycrypto deps). However, I am able to install Fabric in my system-wide site-packages, no problem. I have installed Django into the project's virtualenv and I am able to use all the " python manage.py" commands easily when I activate the virtualenv with the "VIRTUALENV\Scripts\activate.bat" script. I have a fabric tasks file (fabfile.py) in my project that provides tasks for setup, test, deploy, etc. Some of the tasks in my fabfile need to access the settings of my django project through "from django.conf import settings". Since the only usable Fabric install I have is in my system-wide site-packages, I need to activate the virtualenv within my fabfile so django becomes available. To do this, I use the "activate_this" module of the project's virtualenv in order to have access to the project settings and such. Using "print sys.path" before and after I execute activate_this.py, I can tell the python path changes to point to the virtualenv for the project. However, I still cannot import django.conf.settings. I have been able to successfully do this on *nix (Ubuntu and CentOS) and in Cygwin. Do you use this setup/workflow on Windows? If so Can you help me figure out why this wont work on Windows or provide any tips and tricks to get around this issue? Thanks and Cheers. REF: http://virtualenv.openplans.org/#id9 | Using Virtualenv without bin/python Local development environment: Python 2.5.4 Virtualenv 1.4.6 Fabric 0.9.0 Pip 0.6.1 Django 1.1.1 Windows XP (SP3)

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