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• Python-MySQLdb problem: wrong ELF class: ELFCLASS32

  - by jsalonen

  As part of trying out django CMS (http://www.django-cms.org/), I'm struggling with getting Python-MySQLdb to work (http://pypi.python.org/pypi/MySQL-python/). I have installed Django CMS and all of its dependencies (Python 2.5, Django, django-south, MySQL server) I'm trying out the example code within Django CMS code with MySQL as chosen database type When I execute python manage.py syncdb, the following error occurs: django.core.exceptions.ImproperlyConfigured: Error loading MySQLdb module: /root/.python-eggs/MySQL_python-1.2.3c1-py2.5-linux-i686.egg-tmp/_mysql.so: wrong ELF class: ELFCLASS32 I have been able to trace the problem specifically to python-mySQLdb (as also visible in the stack trace). Other than that, I am completely puzzled. I don't have a clue what ELFCLASS32 means, or what ELF class is anyway. I suspect that this error could have something to do with the fact that I am running 64-bit version of Debian 5 (on a VPS). Any good ideas how to troubleshoot?

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• django app using amazon aws s3 storage in stead of DB?

  - by farble1670

  new to python here so bear with me ... i'm looking at django for a rapid prototype to a photo sharing app with an amazon aws s3 storage back end. however, as far as i can tell, django is tailored toward the typical database MVC type of pattern. is there a way to for example provide a custom django model implementation that talks to s3 in stead of a DB? a custom DB engine? would either of these be practical, or am i looking in the wrong direction? thanks.

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• Storing hierarchical (parent/child) data in Python/Django: MPTT alternative?

  - by Parand

  I'm looking for a good way to store and use hierarchical (parent/child) data in Django. I've been using django-mptt, but it seems entirely incompatible with my brain - I end up with non-obvious bugs in non-obvious places, mostly when moving things around in the tree: I end up with inconsistent state, where a node and its parent will disagree on their relationship. My needs are simple: Given a node: find its root find its ancestors find its descendants With a tree: easily move nodes (ie. change parent) My trees will be smallish (at most 10k nodes over 20 levels, generally much much smaller, say 10 nodes with 1 or 2 levels). I have to think there has to be an easier way to do trees in python/django. Are there other approaches that do a better job of maintaining consistency?

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• Actionscript multiple file upload, with parameter passing is not working

  - by Metropolis

  Hey everyone, First off, I am very bad at flash/actionscript, it is not my main programming language. I have created my own file upload flash app that has been working great for me up until this point. It uses PHP to upload the files and sends back a status message which gets displayed in a status box to the user. Now I have run into a situation where I need the HTML to pass a parameter to the Actionscript, and then to the PHP file using POST. I have tried to set this up just like adobe has it on http://livedocs.adobe.com/flex/3/html/help.html?content=17_Networking_and_communications_7.html without success. Here is my Actionscript code import fl.controls.TextArea; //Set filters var imageTypes:FileFilter = new FileFilter("Images (*.jpg, *.jpeg, *.gif, *.png)", "*.jpg; *.jpeg; *.gif; *.png"); var textTypes:FileFilter = new FileFilter("Documents (*.txt, *.rtf, *.pdf, *.doc)", "*.txt; *.rtf; *.pdf; *.doc"); var allTypes:Array = new Array(textTypes, imageTypes); var fileRefList:FileReferenceList = new FileReferenceList(); //Add event listeners for its various fileRefList functions below upload_buttn.addEventListener(MouseEvent.CLICK, browseBox); fileRefList.addEventListener(Event.SELECT, selectHandler); function browseBox(event:MouseEvent):void { fileRefList.browse(allTypes); } function selectHandler(event:Event):void { var phpRequest:URLRequest = new URLRequest("ajax/upload.ajax.php"); var flashVars:URLVariables = objectToURLVariables(this.root.loaderInfo); phpRequest.method = URLRequestMethod.POST; phpRequest.data = flashVars; var file:FileReference; var files:FileReferenceList = FileReferenceList(event.target); var selectedFileArray:Array = files.fileList; var listener:Object = new Object(); for (var i:uint = 0; i < selectedFileArray.length; i++) { file = FileReference(selectedFileArray[i]); try { file.addEventListener(DataEvent.UPLOAD_COMPLETE_DATA, phpResponse); file.upload(phpRequest); } catch (error:Error) { status_txt.text = file.name + " Was not uploaded correctly (" + error.message + ")"; } } } function phpResponse(event:DataEvent):void { var file:FileReference = FileReference(event.target); status_txt.htmlText += event.data; } function objectToURLVariables(parameters:Object):URLVariables { var paramsToSend:URLVariables = new URLVariables(); for(var i:String in parameters) { if(i!=null) { if(parameters[i] is Array) paramsToSend[i] = parameters[i]; else paramsToSend[i] = parameters[i].toString(); } } return paramsToSend; } The flashVars variable is the one that should contain the values from the HTML file. But whenever I run the program and output the variables in the PHP file I receive the following. //Using this command on the PHP page print_r($_POST); //I get this for output Array ( [Filename] => testfile.txt [Upload] => Submit Query ) Its almost like the parameters are getting over written or are just not working at all. Thanks for any help, Metropolis

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• Right way to have a thread in parallel to django project on wsgi.

  - by Enrico Carlesso

  Hi guys. I'm writing a django project, and I need to have a parallel thread which performs certain tasks. The project will be deployed in Apache2.2 with mod_wsgi. Actually my implementation consists on a thread with a while True - Sleep which is called from my django.wsgi file. Is this implementation correct? Two problems raises: does django.wsgi get called only once? Will I have just that instance of the thread running? And second, I need to "manually" visit at least a page to have the Thread run. Is there a workaround? Does anyone has some hints on better solutions? Thanks in advance.

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• Django admin panel doesn't work after modify default user model.

  - by damienix

  I was trying to extend user profile. I founded a few solutions, but the most recommended was to create new user class containing foreign key to original django.contrib.auth.models.User class. I did it with this so i have in models.py: class UserProfile(models.Model): user = models.ForeignKey(User, unique=True) website_url = models.URLField(verify_exists=False) and in my admin.py from django.contrib import admin from someapp.models import * from django.contrib.auth.admin import UserAdmin # Define an inline admin descriptor for UserProfile model class UserProfileInline(admin.TabularInline): model = UserProfile fk_name = 'user' max_num = 1 # Define a new UserAdmin class class MyUserAdmin(UserAdmin): inlines = [UserProfileInline, ] # Re-register UserAdmin admin.site.unregister(User) admin.site.register(User, MyUserAdmin) And now when I'm trying to create/edit user in admin panel i have an error: "Unknown column 'content_userprofile.id' in 'field list'" where content is my appname. I was trying to add line AUTH_PROFILE_MODULE = 'content.UserProfile' to my settings.py but with no effect. How to tell panel admin to know how to correctly display fields in user form?

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• Iterating through folders and files in batch file?

  - by Will Marcouiller

  Here's my situation. A project has as objective to migrate some attachments to another system. These attachments will be located to a parent folder, let's say "Folder 0" (see this question's diagram for better understanding), and they will be zipped/compressed. I want my batch script to be called like so: BatchScript.bat "c:\temp\usd\Folder 0" I'm using 7za.exe as the command line extraction tool. What I want my batch script to do is to iterate through the "Folder 0"'s subfolders, and extract all of the containing ZIP files into their respective folder. It is obligatory that the files extracted are in the same folder as their respective ZIP files. So, files contained in "File 1.zip" are needed in "Folder 1" and so forth. I have read about the FOR...DO command on Windows XP Professional Product Documentation - Using Batch Files. Here's my script: @ECHO OFF FOR /D %folder IN (%%rootFolderCmdLnParam) DO FOR %zippedFile IN (*.zip) DO 7za.exe e %zippedFile I guess that I would also need to change the actual directory before calling 7za.exe e %zippedFile for file extraction, but I can't figure out how in this batch file (through I know how in command line, and even if I know it is the same instruction "cd"). Anyone's help is gratefully appreciated.

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• Windows file association for README, INSTALL, LICENSE and the like [closed]

  - by Lumi

  Possible Duplicate: How to set the default program for opening files without an extension in Windows? Many files originating in the UNIX world come without file extension. Popular examples include README, INSTALL, LICENSE. We know for a fact that these are text files. It is therefore a bit disappointing not to be able to just double-click them open in Explorer and see them in Notepad (actually, Notepad2 because of the UNIX line endings which silly Microsoft Notepad doesn't render correctly). Does anyone know of a way to create a file association for, say, README files without extension? This could then be replicated to cover the most frequently occurring file types, and then double-clicking them open would work. Update (Sort of in response to all your comments.) Thanks, folks, your comments and answers have helped me. @Indrek, yes, I was under the assumption that you could somehow create an association for just README or Makefile, and couldn't do so for files without extension. Turns out the contrary is true, and yes, that is a workaround that neatly solves the issue. Ultimately, I just want to be able to double-click to open a README or Makefile, that's all. @Sampo, the SendMe trick is also useful, although usability is not as great as a straight double-click. (I'm really lazy sometimes.) Turns out the following trick using ftype and ftype from an Administrator prompt does the double-click enabling job: assoc .=no_ext ftype no_ext=%SystemRoot%\system32\NOTEPAD.EXE %1 :: You can see it created some entries in the registry: reg query hkcr\no_ext /s reg query hkcr\. /s

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• Moving hidden files/folders with the command-line or batch-file

  - by Synetech

  Question Does anyone know of a way to move files and folders that have the hidden, system, or read-only attribute set from the command-line or a batch file? (No, stripping the attributes first is not an option since there is no practical way to know which attributes were set in order to re-set them after the move.) (Failed) Attempts Using the basic move command does not work with items with the hidden or system attribute set and for some reason, it does not have switches to specify attributes like the dir and del commands do. I tried using a utility I wrote that uses the shell’s file operation function, but that requires using start /w to prevent the batch file from running on ahead, and it complains about long-filename support for some reason. I tried using robocopy, but it first copies the files and then deletes the originals instead of simply moving the source (which results in a frustrating delay, even with the excessive output redirected to nul). (Surprisingly it seems that few people have ever needed to move hidden files from the command-line. All I could find was this one person who abandoned the attempt.)

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• Windows network: deny file access to another user if file is currently open

  - by Steve

  Some users on my network are having difficulties saving files, because the file is open elsewhere. Let's say Lemuel wants to edit a file, but Bernice in the next office over is working on it. Lemuel opens, edits, and tries to save, but then gets a "no write access" error. Bernice chortles with glee (since earlier that week Lemuel stole her sandwich). Unfortunately, various softwares will not warn the user that they have opened a read-only file. Is there a way (in Windows) to limit file access to ONE user only, i.e. 777 for the first user to open the file, and 000 for all users after that? (Sorry for the Linux terminology but it gets my point across).

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• Server 2008 R2 file access permissions

  - by Napster100

  I'm finding it awkward to sort out permissions for file sharing and access on my LAN. I've created an account on the server node (as a normal user) and shared a drive that has 2 folders at the root, one is for personal file storage and the other shared files, if I connect to the shared area from a workstation running windows 7 and log-in using the account I created on the server, I can look through directories but can't look in some (which I wanted as I changed the permissions for that to happen), but my problem is although the permissions are set for this user account to have full control of the specific folder I can't create a folder in that area or upload files to that folder. Could someone explain why this is? Thanks in advanced

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• Java - Display % of upload done

  - by tr-raziel

  I have a java applet for uploading files to server. I want to display the % of data sent but when I use ObjectOutputStream.write() it just writes to the buffer, does not wait until the data has actually been sent. How can I achieve this. Perhaps I need to use thread synchronization or something. Any clues would be most helpful. This is the code I'm using right now: try{ for(File file : ficheiros){ FileInputStream stream = new FileInputStream (file); int bytesRead1 = 0;; int off1 = 0; int len1 = 100000; if(file.length() < 100000) len1 = new Long(file.length()).intValue(); byte[] bytes1 = new byte[len1]; while (off1 < file.length()) { bytes1 = new byte[len1]; if((file.length() - off1) < len1){ len1 = (new Long(file.length()).intValue() - off1); bytes1 = new byte[len1]; } if((bytesRead1 = stream.read(bytes1)) != -1){ //I want this to block until all data has been sent outputToServlet.write(bytes1, 0, bytesRead1 ); System.out.println("off1: " + off1); off1 = off1 + len1; outputToServlet.flush(); } sent += len1; if(sent>totalLength) sent = (int)totalLength; updateFeedback(sent,totalLength,false);//calls method to display % } updateFeedback(-1,-1,true); } }catch(Exception e){ e.printStackTrace(); } Thanks

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• php file upload problem [closed]

  - by newcomer

  This code works properly in my localhost. I am using xampp 1.7.3. but when I put it in the live server it shows Possible file upload attack!. 'upload/' is the folder under 'public_html' folder on the server. I can upload files via other script in that directory. <?php $uploaddir = '/upload/';//I used C:/xampp/htdocs/upload/ in localhost. is it correct here? $uploadfile = $uploaddir . basename($_FILES['file_0']['name']); echo '<pre>'; if (move_uploaded_file($_FILES['file_0']['tmp_name'], $uploadfile)) { echo "File is valid, and was successfully uploaded.\\n"; } else { echo "Possible file upload attack!\\n"; } echo 'Here is some more debugging info:'; print_r($_FILES); print "</pre>"; ?>

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• Django - The included urlconf doesn't have any patterns in it

  - by unsorted

  My website, which was working before, suddenly started breaking with the error "ImproperlyConfigured at / The included urlconf resume.urls doesn't have any patterns in it" The project base is called resume. In settings.py I have set ROOT_URLCONF = 'resume.urls' Here's my resume.urls, which sits in the project root directory. from django.conf.urls.defaults import * # Uncomment the next two lines to enable the admin: from django.contrib import admin admin.autodiscover() urlpatterns = patterns('', # Example: # (r'^resume/', include('resume.foo.urls')), # Uncomment the admin/doc line below and add 'django.contrib.admindocs' # to INSTALLED_APPS to enable admin documentation: (r'^admin/doc/', include('django.contrib.admindocs.urls')), # Uncomment the next line to enable the admin: (r'^admin/', include(admin.site.urls)), (r'^accounts/login/$', 'django.contrib.auth.views.login'), #(r'^employer/', include(students.urls)), (r'^ajax/', include('urls.ajax')), (r'^student/', include('students.urls')), (r'^club/(?P<object_id>\d+)/$', 'resume.students.views.club_detail'), (r'^company/(?P<object_id>\d+)/$', 'resume.students.views.company_detail'), (r'^program/(?P<object_id>\d+)/$', 'resume.students.views.program_detail'), (r'^course/(?P<object_id>\d+)/$', 'resume.students.views.course_detail'), (r'^career/(?P<object_id>\d+)/$', 'resume.students.views.career_detail'), (r'^site_media/(?P<path>.*)$', 'django.views.static.serve', {'document_root': 'C:/code/django/resume/media'}), ) Anyone know what's wrong? This is driving me crazy. Thanks,

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• Settings module not found deploying django on a shared server

  - by mcanes

  I'm trying to deploy my django project on a shared hosting as describe here I have my project on /home/user/www/testa I'm using this script #!/usr/bin/python import sys, os sys.path.append("/home/user/bin/python") sys.path.append('/home/user/www/testa') os.chdir("/home/user/www/testa") os.environ['DJANGO_SETTINGS_MODULE'] = "settings.py" from django.core.servers.fastcgi import runfastcgi runfastcgi(method="threaded", daemonize="false") And here's the error I get when trying to run it from shell: WSGIServer: missing FastCGI param REQUEST_METHOD required by WSGI! WSGIServer: missing FastCGI param SERVER_NAME required by WSGI! WSGIServer: missing FastCGI param SERVER_PORT required by WSGI! WSGIServer: missing FastCGI param SERVER_PROTOCOL required by WSGI! Traceback (most recent call last): File "build/bdist.linux-i686/egg/flup/server/fcgi_base.py", line 558, in run File "build/bdist.linux-i686/egg/flup/server/fcgi_base.py", line 1118, in handler File "/home/user/lib/python2.4/site-packages/django/core/handlers/wsgi.py", line 230, in __call__ self.load_middleware() File "/home/user/lib/python2.4/site-packages/django/core/handlers/base.py", line 33, in load_middleware for middleware_path in settings.MIDDLEWARE_CLASSES: File "/home/user/lib/python2.4/site-packages/django/utils/functional.py", line 269, in __getattr__ self._setup() File "/home/usr/lib/python2.4/site-packages/django/conf/__init__.py", line 40, in _setup self._wrapped = Settings(settings_module) File "/home/user/lib/python2.4/site-packages/django/conf/__init__.py", line 75, in __init__ raise ImportError, "Could not import settings '%s' (Is it on sys.path? Does it have syntax errors?): %s" % (self.SETTINGS_MODULE, e) ImportError: Could not import settings 'settings.py' (Is it on sys.path? Does it have syntax errors?): No module named settings.py Content-Type: text/html Unhandled Exception Unhandled Exception An unhandled exception was thrown by the application. What am I doing wrong? Running the script from the browser just gives me an internal server error.

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• Django generic relation field reports that all() is getting unexpected keyword argument when no args

  - by Joshua

  I have a model which can be attached to to other models. class Attachable(models.Model): content_type = models.ForeignKey(ContentType) object_pk = models.TextField() content_object = generic.GenericForeignKey(ct_field="content_type", fk_field="object_pk") class Meta: abstract = True class Flag(Attachable): user = models.ForeignKey(User) flag = models.SlugField() timestamp = models.DateTimeField() I'm creating a generic relationship to this model in another model. flags = generic.GenericRelation(Flag) I try to get objects from this generic relation like so: self.flags.all() This results in the following exception: >>> obj.flags.all() Traceback (most recent call last): File "<console>", line 1, in <module> File "/usr/local/lib/python2.6/dist-packages/django/db/models/manager.py", line 105, in all return self.get_query_set() File "/usr/local/lib/python2.6/dist-packages/django/contrib/contenttypes/generic.py", line 252, in get_query_set return superclass.get_query_set(self).filter(**query) File "/usr/local/lib/python2.6/dist-packages/django/db/models/query.py", line 498, in filter return self._filter_or_exclude(False, *args, **kwargs) File "/usr/local/lib/python2.6/dist-packages/django/db/models/query.py", line 516, in _filter_or_exclude clone.query.add_q(Q(*args, **kwargs)) File "/usr/local/lib/python2.6/dist-packages/django/db/models/sql/query.py", line 1675, in add_q can_reuse=used_aliases) File "/usr/local/lib/python2.6/dist-packages/django/db/models/sql/query.py", line 1569, in add_filter negate=negate, process_extras=process_extras) File "/usr/local/lib/python2.6/dist-packages/django/db/models/sql/query.py", line 1737, in setup_joins "Choices are: %s" % (name, ", ".join(names))) FieldError: Cannot resolve keyword 'object_id' into field. Choices are: content_type, flag, id, nestablecomment, object_pk, timestamp, user >>> obj.flags.all(object_pk=obj.pk) Traceback (most recent call last): File "<console>", line 1, in <module> TypeError: all() got an unexpected keyword argument 'object_pk' What have I done wrong?

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• Installing django on dreamhost (help a newb out)

  - by augustfirst

  I'm trying to get django running on my dreahost account. I've been trying to sort of use two tutorials at once: the one on the dreamhost wiki and the one in the django book. I installed django using the script on the wiki page, but I ran into trouble immediately while trying to work through the django book. It says: To start the server, change into your project directory (cd mysite), if you haven’t already, and run this command: python manage.py runserver This launches the server locally, on port 8000, accessible only to connections from your own computer. Now that it’s running, visit 127.0.0.1:8000 with your Web browser. You’ll see a “Welcome to Django” page shaded in a pleasant pastel blue. It worked! Those instructions seem to assume that you're developing locally, not on a shared server. Where the heck am I supposed to look for the "Welcome to Django" page after starting the server? In my webroot? No dice. Anyway, I tried to blunder ahead through the django book to its hello world tutorial (chapter 3). But once I've edited the view file and the URLconf, I don't get a nice clean "hello world" text. Instead (as you can see) I get an "import error". Any help would be greatly appreciated.

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• Need help with Django tutorial

  - by Nai

  I'm doing the Django tutorial here: http://docs.djangoproject.com/en/1.2/intro/tutorial03/ My TEMPLATE_DIRS in the settings.py looks like this: TEMPLATE_DIRS = ( "/webapp2/templates/" "/webapp2/templates/polls" # Put strings here, like "/home/html/django_templates" or "C:/www/django/templates". # Always use forward slashes, even on Windows. # Don't forget to use absolute paths, not relative paths. ) My urls.py looks like this: from django.conf.urls.defaults import * from django.contrib import admin admin.autodiscover() urlpatterns = patterns('', (r'^polls/$', 'polls.views.index'), (r'^polls/(?P<poll_id>\d+)/$', 'polls.views.detail'), (r'^polls/(?P<poll_id>\d+)/results/$', 'polls.views.results'), (r'^polls/(?P<poll_id>\d+)/vote/$', 'polls.views.vote'), (r'^admin/', include(admin.site.urls)), ) My views.py looks like this: from django.template import Context, loader from polls.models import Poll from django.http import HttpResponse def index(request): latest_poll_list = Poll.objects.all().order_by('-pub_date')[:5] t = loader.get_template('c:/webapp2/templates/polls/index.html') c = Context({ 'latest_poll_list': latest_poll_list, }) return HttpResponse(t.render(c)) I think I am getting the path of my template wrong because when I simplify the views.py code to something like this, I am able to load the page. from django.http import HttpResponse def index(request): return HttpResponse("Hello, world. You're at the poll index.") My index template file is located at C:/webapp2/templates/polls/index.html. What am I doing wrong?

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• Upload progress meter needed - PHP

  - by Pawan Kumar

  Hello everyone I have implemented Amazon S3 on my website to upload video. But i want to include upload progress meter in my site to show the status of how much percent, file has been uploaded. If any one have such script please replay me.

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• Upload progress meter needed

  - by Pawan Kumar

  Hello everyone I have implemented Amazon S3 on my website to upload video. But i want to include upload progress meter in my site to show the status of how may percent file has been uploaded. If any one have such script please replay me.

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• Rails upload to s3 performance issue

  - by Denis

  Hello, I'm building an app to store files on my s3 account. I use Rails 3.0.0beta A lot of files can be uploaded at the same time, and the cost (from a performance point of view) of an upload is quite heavy, my app will be busy handling uploads all the time! Maybe a solution is to upload directly to s3, but I still need a submit to my app, at least to store the file's name. I'm wondering what is the best solution?

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• Aborting upload from a servlet

  - by Zizzencs

  I'd like to limit the size of the file that can be uploaded to an application. To achieve this, I'd like to abort the upload process from the server side when the size of the file being uploaded exceeds a limit. Is there a way to abort an upload process from the server side without waiting the HTTP request to finish?

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• How do we get sharepoint upload limit using web services

  - by Royson

  In my application before uploading i have to check whether file size is greater than upload limit. How do we get share point file upload limit using web services in C#??

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• How to upload file outside web server?

  - by Stardust

  I want to upload files outisde web server like in d d drive into servlets, I but I'm not able to upload them. What I have to do to make functionality like this enable in Tomcat 6.0?

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• How can I upload a file from an iPad to a web server (upload servlet)

  - by Chen Harel

  Hi, I'm writing an iPad application that mimics a flash website I built. The site uses Flash file uploader to upload files from the user's filesystem to my tomcat server. I understand that an iPad application can sync files via iTunes to a Documents folder. Given access to those files, how can I invoke a file upload of a selected file to the web.

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