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• Django: TypeError: 'str' object is not callable, referer: http://xxx

  - by user705415

  I've been wondering why when I set the settings.py of my django project 'arvindemo' debug = Flase and deploy it on Apache with mod_wsgi, I got the 500 Internal Server Error. Env: Django 1.4.0 Python 2.7.2 mod_wsgi 2.8 OS centOS Here is the recap: Visit the homepage, go to sub page A/B/C/D, and fill some forms, then submit it to the Apache server. Once click 'submit' button, I will get the '500 Internal Server Error', and the error_log listed below(Traceback): [Tue Apr 10 10:07:20 2012] [error] [client 122.198.133.250] Traceback (most recent call last): [Tue Apr 10 10:07:20 2012] [error] [client 122.198.133.250] File "/opt/python2.7/lib/python2.7/site-packages/django/core/handlers/wsgi.py", line 241, in __call__ [Tue Apr 10 10:07:20 2012] [error] [client 122.198.133.250] response = self.get_response(request) [Tue Apr 10 10:07:20 2012] [error] [client 122.198.133.250] File "/opt/python2.7/lib/python2.7/site-packages/django/core/handlers/base.py", line 179, in get_response [Tue Apr 10 10:07:20 2012] [error] [client 122.198.133.250] response = self.handle_uncaught_exception(request, resolver, sys.exc_info()) [Tue Apr 10 10:07:20 2012] [error] [client 122.198.133.250] File "/opt/python2.7/lib/python2.7/site-packages/django/core/handlers/base.py", line 224, in handle_uncaught_exception [Tue Apr 10 10:07:20 2012] [error] [client 122.198.133.250] if resolver.urlconf_module is None: [Tue Apr 10 10:07:20 2012] [error] [client 122.198.133.250] File "/opt/python2.7/lib/python2.7/site-packages/django/core/urlresolvers.py", line 323, in urlconf_module [Tue Apr 10 10:07:20 2012] [error] [client 122.198.133.250] self._urlconf_module = import_module(self.urlconf_name) [Tue Apr 10 10:07:20 2012] [error] [client 122.198.133.250] File "/opt/python2.7/lib/python2.7/site-packages/django/utils/importlib.py", line 35, in import_module [Tue Apr 10 10:07:20 2012] [error] [client 122.198.133.250] __import__(name) [Tue Apr 10 10:07:20 2012] [error] [client 122.198.133.250] File "/opt/web/django/arvindemo/arvindemo/../arvindemo/urls.py", line 23, in <module> [Tue Apr 10 10:07:20 2012] [error] [client 122.198.133.250] url(r'^submitPage$', name=submitPage), [Tue Apr 10 10:07:20 2012] [error] [client 122.198.133.250] TypeError: url() takes at least 2 arguments (2 given) When using django runserver, I set arvindemo.settings debug = True, everything is OK. But things changed once I set debug = Flase. Here is my views.py from django.http import HttpResponseRedirect from django.http import HttpResponse, HttpResponseServerError from django.shortcuts import render_to_response import datetime, string from user_info.models import * from django.template import Context, loader, RequestContext import settings def hello(request): return HttpResponse("hello girl") def helpPage(request): return render_to_response('kktHelp.html') def server_error(request, template_name='500.html'): return render_to_response(template_name, context_instance = RequestContext(request) ) def page404(request): return render_to_response('404.html') def submitPage(request): post = request.POST Mall = 'goodsName' Contest = 'ojs' Presentation = 'addr' WeatherReport = 'city' Habit = 'task' if Mall in post: return submitMall(request) elif Contest in post: return submitContest(request) elif Presentation in post: return submitPresentation(request) elif Habit in post: return submitHabit(request) elif WeatherReport in post: return submitWeather(request) else: return HttpResponse(request.POST) return HttpResponseRedirect('404') def submitXXX(): ..... def xxxx(): .... Here comes the urls.py from django.conf.urls import patterns, include, url from views import * from django.conf import settings handler500 = 'server_error' urlpatterns = patterns('', url(r'^hello/$', hello), # hello world url(r'^$', homePage), url(r'^time/$', getTime), url(r'^time/plus/(\d{1,2})/$', hoursAhead), url(r'^Ttime/$', templateGetTime), url(r'^Mall$', templateMall), url(r'^Contest$', templateContest), url(r'^Presentation$', templatePresentation), url(r'^Habit$', templateHabit), url(r'^Weather$', templateWeather), url(r'^Help$', helpPage), url(r'^404$', page404), url(r'^500$', server_error), url(r'^submitPage$', submitPage), url(r'^submitMall$', submitMall), url(r'^submitContest$', submitContest), url(r'^submitPresentation$', submitPresentation), url(r'^submitHabit$', submitHabit), url(r'^submitWeather$', submitWeather), url(r'^terms$', terms), url(r'^privacy$', privacy), url(r'^thanks$', thanks), url(r'^about$', about), url(r'^static/(?P<path>.*)$','django.views.static.serve',{'document_root':settings.STATICFILES_DIRS}), ) I'm sure there is no syntax error in my django project,cause when I use django runserver, everything is fine. Anyone can help ? Best regards

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• How to render a POST and make it show up on another page

  - by stack5914

  I'm trying to create a marketplace website similar to craigslist. I created a form according to the Django tutorial "Working with forms", but I don't know how to render information I got from the POST forms. I want to make information(subject,price...etc) that I got from POST show up on another page like this. http://bakersfield.craigslist.org/atq/3375938126.html and, I want the "Subject"(please look at form.py) of this product(eg.1960 French Chair) to show up on another page like this. http://bakersfield.craigslist.org/ata/ } Can I get some advice to handle submitted information? Here's present codes. I'll appreciate all your answers and helps. <-! Here's my codes -- ?forms.py from django import forms class SellForm(forms.Form): subject = forms.CharField(max_length=100) price = forms.CharField(max_length=100) condition = forms.CharField(max_length=100) email = forms.EmailField() body = forms.TextField() ?views.py from django.shortcuts import render, render_to_response from django.http import HttpResponseRedirect from site1.forms import SellForm def sell(request): if request.method =="POST": form =SellForm(request.POST) if form.is_valid(): subject = form.cleaned_data['subject'] price = form.cleaned_data['price'] condition = form.cleaned_data['condition'] email = form.cleaned_data['email'] body = form.cleaned_data['body'] return HttpResponseRedirect('/books/') else: form=SellForm() render(request, 'sell.html',{'form':form,}) ?urls.py from django.conf.urls import patterns, include, url from django.contrib import admin admin.autodiscover() urlpatterns = patterns('', url(r'^sechand/$','site1.views.sell'), url(r'^admin/', include(admin.site.urls)), ) ?sell.html <form action = "/sell/" method = "post">{% csrf_token%} {{ form.as_p }} <input type = "submit" value="Submit" /> </form>

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• How to stop Django from adding extra html elements to rendered widgets.

  - by stinkypyper

  I have a Django radio button group that renders to HTML as follows: <ul> <li><label for="id_package_id_0"><input type="radio" id="id_package_id_0" value="1" name="package_id" /> Test 256</label></li> <li><label for="id_package_id_1"><input type="radio" id="id_package_id_1" value="2" name="package_id" /> Test 384</label></li> <li><label for="id_package_id_2"><input type="radio" id="id_package_id_2" value="3" name="package_id" /> Test 512</label></li> <li><label for="id_package_id_3"><input type="radio" id="id_package_id_3" value="4" name="package_id" /> Test 768</label></li> <li><label for="id_package_id_4"><input type="radio" id="id_package_id_4" value="5" name="package_id" /> Test 1024</label></li> </ul> I need it to render without being a list. I am a aware of form.as_p, form.as_table, and form.as_ul. They will not help me as they continue to add extra HTML tags. As well, I am not using the form object in it's absolute entirety, just for validation. I am doing a custom template for the form already, but wish to continue to the radio widget.

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• Django debug error

  - by Hulk

  I have the following in my model: class info(models.Model): add = models.CharField(max_length=255) name = models.CharField(max_length=255) An in the views when i say info_l = info.objects.filter(id=1) logging.debug(info_l.name) i get an error saying name doesnt exist at debug statement. 'QuerySet' object has no attribute 'name' 1.How can this be resolved. 2.Also how to query for only one field instead of selecting all like select name from info.

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• Django, ModelForms, User and UserProfile - not hashing password

  - by IvanBernat

  I'm trying to setup a User - UserProfile relationship, display the form and save the data. When submitted, the data is saved, except the password field doesn't get hashed. Additionally, how can I remove the help_text from the username and password (inherited from the User model)? Full code is below, excuse me if it's too long. Models.py USER_IS_CHOICES = ( ('u', 'Choice A'), ('p', 'Choice B'), ('n', 'Ninja'), ) class UserProfile(models.Model): user = models.ForeignKey(User, unique=True) user_is = models.CharField(max_length=1, choices=USER_IS_CHOICES) Forms.py class UserForm(forms.ModelForm): class Meta: model = User fields = ["first_name", "last_name", "username", "email", "password"] def clean_username(self): username = self.cleaned_data['username'] if not re.search(r'^\w+$', username): raise forms.ValidationError('Username can contain only alphanumeric characters') try: User.objects.get(username=username) except ObjectDoesNotExist: return username raise forms.ValidationError('Username is already taken') class UserProfileForm(forms.ModelForm): class Meta: model = UserProfile fields = ['user_is'] Views.py if request.method == 'POST': uform = UserForm(request.POST) pform = UserProfileForm(request.POST) if uform.is_valid() and pform.is_valid(): user = uform.save() profile = pform.save(commit = False) profile.user = user profile.save() return HttpResponseRedirect('/') else: uform = UserForm() pform = UserProfileForm() variables = RequestContext(request, { 'uform':uform, 'pform':pform }) return render_to_response('registration/register.html', variables)

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• django admin site make CharField a PasswordInput

  - by Paul

  I have a Django site in which the site admin inputs their Twitter Username/Password in order to use the Twitter API. The Model is set up like this: class TwitterUser(models.Model): screen_name = models.CharField(max_length=100) password = models.CharField(max_length=255) def __unicode__(self): return self.screen_name I need the Admin site to display the password field as a password input, but can't seem to figure out how to do it. I have tried using a ModelAdmin class, a ModelAdmin with a ModelForm, but can't seem to figure out how to make django display that form as a password input...

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• Django Model Formset Pre-Filled Value Problem

  - by user552377

  Hi, i'm trying to use model formsets with Django. When i load forms template, i see that it's filled-up with previous values. Is there a caching mechanism that i should stop, or what? Thanks for your help, here is my code: models.py class FooModel( models.Model ): a_field = models.FloatField() b_field = models.FloatField() def __unicode__( self ): return self.a_field forms.py from django.forms.models import modelformset_factory FooFormSet = modelformset_factory(FooModel) views.py def foo_func(request): if request.method == 'POST': formset = FooFormSet(request.POST, request.FILES, prefix='foo_prefix' ) if formset.is_valid(): formset.save() return HttpResponseRedirect( '/true/' ) else: return HttpResponseRedirect( '/false/' ) else: formset = FooFormSet(prefix='foo_prefix') variables = RequestContext( request , { 'formset':formset , } ) return render_to_response('footemplate.html' , variables ) template: <form method="post" action="."> {% csrf_token %} <input type="submit" value="Submit" /> <table id="FormsetTable" border="0" cellpadding="0" cellspacing="0"> <tbody> {% for form in formset.forms %} <tr> <td>{{ form.a_field }}</td> <td>{{ form.b_field }}</td> </tr> {% endfor %} </tbody> </table> {{ formset.management_form }} </form>

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• Django admin - remove field if editing an object

  - by John McCollum

  I have a model which is accessible through the Django admin area, something like the following: # model class Foo(models.Model): field_a = models.CharField(max_length=100) field_b = models.CharField(max_length=100) # admin.py class FooAdmin(admin.ModelAdmin): pass Let's say that I want to show field_a and field_b if the user is adding an object, but only field_a if the user is editing an object. Is there a simple way to do this, perhaps using the fields attribute? If if comes to it, I could hack a JavaScript solution, but it doesn't feel right to do that at all!

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• Sorting related objects in the Django Admin form interface

  - by Carver

  I am looking to sort the related objects that show up when editing an object using the admin form. So for example, I would like to take the following object: class Person(models.Model): first_name = models.CharField( ... ) last_name = models.CharField( ... ) hero = models.ForeignKey( 'self', null=True, blank=True ) and edit the first name, last name and hero using the admin interface. I want to sort the objects as they show up in the drop down by last name, first name (ascending). How do I do that? Context I'm using Django v1.1. I started by looking for help in the django admin docs, but didn't find the solution As you can see in the example, the foreign key is pointing to itself, but I expect it would be the same as pointing to a different model object. Bonus points for being able to filter the related objects, too (eg~ only allow selecting a hero with the same first name)

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• reverse many to many fields in Django + count them

  - by cleliodpaula

  I'm trying to figure out how to solve this class Item(models.Model): type = models.ForeignKey(Type) name = models.CharField(max_lenght = 10) ... class List(models.Model): items = models.ManyToManyField(Item) ... I want to count how many an Item appears in another Lists, and show on template. view def items_by_list(request, id_): list = List.objects.get(id = id_) qr = list.items.all() #NOT TESTED num = [] i = 0 for item in qr: num[i] = List.objects.filter(items__id = item__id ).count() #FINISH NOT TESTED c = {} c.update(csrf(request)) c = {'request':request, 'list' : qr, 'num' : num} return render_to_response('items_by_list.html', c, context_instance=RequestContext(request)) template {% for dia in list %} <div class="span4" > <div> <h6 style="color: #9937d8">{{item.type.description}}</h6> <small style="color: #b2e300">{{ item.name }}</small> <small style="color: #b2e300">{{COUNT HOW MANY TIMES THE ITEM APPEAR ON OTHER LISTS}}</small> </div> {% endfor %} This seems to be easy, but I could not implement yet. If anyone has some glue to me, please help me. Thanks in advance.

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• Django ImageField issue with JPEG's

  - by Kieran Lynn

  I am having a major issue with PIL (Python Image Library) in Django and have jumpped through a lot of hoops and have thus far not been able to figure out what the root of the issue is. The problem essentially breaks down to not being able to upload JPEG images through the ImageField in the Django admin. But the issue is not as simple as installing libjpeg. First, I installed PIL (through Buildout) and realized once it was installed that I had not installed libjpeg because JPEG support was not available. Having not setup the server myself, I just assumed that it was not installed and I compiled libjpeg 8 from the source. This ended up in my /usr/local/lib/ directory. I cleared out my Buildout files and rebuilt everything. This time when PIL compiled I had JPEG support. But I went to the Django Admin and tried to upload a JPEG though an ImageField with no luck. I got the "Upload a valid image. The file you uploaded was either not an image or a corrupted image" error. Just as a test I opened up a the Djano shell and ran the following: > import Image > i = Image.open( "/absolute_path/file.jpg" ) > print i <JpegImagePlugin.JpegImageFile image mode=RGB size=940x375 at 0x7F908C529BD8> This runs with no errors and shows that PIL is able to open JPEG's. After doing some reading, I come across this thread: Is it possible to control which libraries apache uses? Looks like PHP also uses libjpeg and is loading before Django, and therefor loading libjpeg 6.2 before. This is show when using lsof: COMMAND PID USER FD TYPE DEVICE SIZE/OFF NODE NAME apache2 2561 www-data mem REG 202,1 146032 639276 /usr/lib/libjpeg.so.62.0.0 So my thought is that I should be using libjpeg 6.2. So I removed libjpeg located in my /usr/local/lib directory. After rereading the PIL installation instructions, I realized that I might not have the dev/header files for libjpeg that PIL needs. So I also uninstalled libjpeg using the aptitude uninstaller (sudo aptitude remove libjpeg62). Then to ensure that I got the header files that PIL needed I installed libjpeg using aptitude: (sudo aptget install libjpeg62-dev). From here I cleaned out my Buildout directory, and reran Buildout, which in turn reinstalled PIL. Once again, I have JPEG support, now using the libjpeg62. So I go to test in the Django Admin. Still no JPEG support. So I wanted to test JPEG support in general and see if the exception was not handled, what kind of error it would throw. So in my homepage view I added the following code to open a JPEG image: import Image i = Image.open( "/absolute_path/file.jpg" ) v = i.verify() Then I pass i to the HTML view just to easily see the output. I deploy these changes to the server and restart. I am surprised not to see an error and get the following output: {{ i }} - <JpegImagePlugin.JpegImageFile image mode=RGB size=940x375 at 0x7F908C529BD8> {{ v }} - None So at this point I am really confused: Why can I successfully open a JPEG while the admin cannot? Am I missing something, is this not an issue with libjpeg? If not an issue with libjpeg, why can I upload a PNG with no issues? Any help would be much appreciated, I have been on this for 2 days debugging with no luck. Setup: 1. Rackspace Cloud Server 2. Ubuntu 10.04 3. Django 1.2.3 (Installed though Buildout) 4. PIL 1.1.7 (Installed though Buildout) 5. libjpeg 6.2 (installed through aptitude (sudo aptget install libjpeg62-dev)

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• jqgrid and django models

  - by Andrew Gee

  Hi, I have the following models class Employee(Person): job = model.Charfield(max_length=200) class Address(models.Model): street = models.CharField(max_length=200) city = models.CharField(max_length=200) class EmpAddress(Address): date_occupied = models.DateField() date_vacated = models.DateField() employee = models.ForeignKey() When I build a json data structure for an EmpAddress object using the django serialzer it does not include the inherited fields only the EmpAddress fields. I know the fields are available in the object in my view as I can print them but they are not built into the json structure. Does anyone know how to overcome this? Thanks Andrew

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• Django - model.save(commit=False - Is there a way to replicate this?

  - by orokusaki

  I'm wanting to do this: from django.contrib.auth.models import User class PetFrog(models.Model): user = models.OnetoOneField(User) color = models.CharField(max_length=20) def clean(self): if self.color == 'Green': user = User(username='prince') user.save(commit=False) # No commit argument in models.Model.save() like there is in ModelForm.save() user.set_password(self.password) user.save() self.user = user Is there a way to do this creation of a model instance without filling in all the required fields, and then setting them manually before trying to save() for real (which would obviously raise a "Must choose a Password" error)? I need to do this in my model, vs using a ModelForm. If there is another way to do it (while still in clean()), I'm completely open to any suggestions.

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• How do I add a trailing slash for Django MPTT-based categorization app?

  - by Patrick Beeson

  I'm using Django-MPTT to develop a categorization app for my Django project. But I can't seem to get the regex pattern for adding a trailing slash that doesn't also break on child categories. Here's an example URL: http://mydjangoapp.com/categories/parentcat/childcat/ I'd like to be able to use http://mydjangoapp.com/categories/parentcat and have it redirect to the trailing slash version. The same should apply to http://mydjangoapp.com/categories/parentcat/childcat (it should redirect to http://mydjangoapp.com/categories/parentcat/childcat/). Here's my urls.py: from django.conf.urls.defaults import patterns, include, url from django.views.decorators.cache import cache_page from storefront.categories.models import Category from storefront.categories.views import SimpleCategoryView urlpatterns = patterns('', url(r'^(?P<full_slug>[-\w/]+)', cache_page(SimpleCategoryView.as_view(), 60 * 15), name='category_view'), ) And here is my view: from django.core.exceptions import ImproperlyConfigured from django.core.urlresolvers import reverse from django.views.generic import TemplateView, DetailView from django.views.generic.detail import SingleObjectTemplateResponseMixin, SingleObjectMixin from django.utils.translation import ugettext as _ from django.contrib.syndication.views import Feed from storefront.categories.models import Category class SimpleCategoryView(TemplateView): def get_category(self): return Category.objects.get(full_slug=self.kwargs['full_slug']) def get_context_data(self, **kwargs): context = super(SimpleCategoryView, self).get_context_data(**kwargs) context["category"] = self.get_category() return context def get_template_names(self): if self.get_category().template_name: return [self.get_category().template_name] else: return ['categories/category_detail.html'] And finally, my model: from django.db import models from mptt.models import MPTTModel from mptt.fields import TreeForeignKey class CategoryManager(models.Manager): def get(self, **kwargs): defaults = {} defaults.update(kwargs) if 'full_slug' in defaults: if defaults['full_slug'] and defaults['full_slug'][-1] != "/": defaults['full_slug'] += "/" return super(CategoryManager, self).get(**defaults) class Category(MPTTModel): title = models.CharField(max_length=255) description = models.TextField(blank=True, help_text='Please use <a href="http://daringfireball.net/projects/markdown/syntax">Markdown syntax</a> for all text-formatting and links. No HTML is allowed.') slug = models.SlugField(help_text='Prepopulates from title field.') full_slug = models.CharField(max_length=255, blank=True) template_name = models.CharField(max_length=70, blank=True, help_text="Example: 'categories/category_parent.html'. If this isn't provided, the system will use 'categories/category_detail.html'. Use 'categories/category_parent.html' for all parent categories and 'categories/category_child.html' for all child categories.") parent = TreeForeignKey('self', null=True, blank=True, related_name='children') objects = CategoryManager() class Meta: verbose_name = 'category' verbose_name_plural = 'categories' def save(self, *args, **kwargs): orig_full_slug = self.full_slug if self.parent: self.full_slug = "%s%s/" % (self.parent.full_slug, self.slug) else: self.full_slug = "%s/" % self.slug obj = super(Category, self).save(*args, **kwargs) if orig_full_slug != self.full_slug: for child in self.get_children(): child.save() return obj def available_product_set(self): """ Returns available, prioritized products for a category """ from storefront.apparel.models import Product return self.product_set.filter(is_available=True).order_by('-priority') def __unicode__(self): return "%s (%s)" % (self.title, self.full_slug) def get_absolute_url(self): return '/categories/%s' % (self.full_slug)

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• Django admin- constrain visibility of models and model instances based on who logged in

  - by theactiveactor

  Using the django admin, I would like to be able to specify which models a user sees when he logs in. For a stretch goal, for each model types a user can see, I would like to specify a filter to limit which instances of the model the user can see. Could someone please provide a pointer for how to go about achieving this?

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• Muliple Models in a single django ModelForm?

  - by BigJason

  Is it possible to have multiple models included in a single ModelForm in django? I am trying to create a profile edit form. So I need to include some fields from the User model and the UserProfile model. Currently I am using 2 forms like this class UserEditForm(ModelForm): class Meta: model = User fields = ("first_name", "last_name") class UserProfileForm(ModelForm): class Meta: model = UserProfile fields = ("middle_name", "home_phone", "work_phone", "cell_phone") Is there a way to consolidate these into one form or do I just need to create a form and handle the db loading and saving myself?

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• Django + FastCGI - randomly raising OperationalError

  - by ibz

  I'm running a Django application. Had it under Apache + mod_python before, and it was all OK. Switched to Lighttpd + FastCGI. Now I randomly get the following exception (neither the place nor the time where it appears seem to be predictable). Since it's random, and it appears only after switching to FastCGI, I assume it has something to do with some settings. Found a few results when googleing, but they seem to be related to setting maxrequests=1. However, I use the default, which is 0. Any ideas where to look for? PS. I'm using PostgreSQL. Might be related to that as well, since the exception appears when making a database query. Thanks. File "/usr/lib/python2.6/site-packages/django/core/handlers/base.py", line 86, in get_response response = callback(request, *callback_args, **callback_kwargs) File "/usr/lib/python2.6/site-packages/django/contrib/admin/sites.py", line 140, in root if not self.has_permission(request): File "/usr/lib/python2.6/site-packages/django/contrib/admin/sites.py", line 99, in has_permission return request.user.is_authenticated() and request.user.is_staff File "/usr/lib/python2.6/site-packages/django/contrib/auth/middleware.py", line 5, in __get__ request._cached_user = get_user(request) File "/usr/lib/python2.6/site-packages/django/contrib/auth/__init__.py", line 83, in get_user user_id = request.session[SESSION_KEY] File "/usr/lib/python2.6/site-packages/django/contrib/sessions/backends/base.py", line 46, in __getitem__ return self._session[key] File "/usr/lib/python2.6/site-packages/django/contrib/sessions/backends/base.py", line 172, in _get_session self._session_cache = self.load() File "/usr/lib/python2.6/site-packages/django/contrib/sessions/backends/db.py", line 16, in load expire_date__gt=datetime.datetime.now() File "/usr/lib/python2.6/site-packages/django/db/models/manager.py", line 93, in get return self.get_query_set().get(*args, **kwargs) File "/usr/lib/python2.6/site-packages/django/db/models/query.py", line 304, in get num = len(clone) File "/usr/lib/python2.6/site-packages/django/db/models/query.py", line 160, in __len__ self._result_cache = list(self.iterator()) File "/usr/lib/python2.6/site-packages/django/db/models/query.py", line 275, in iterator for row in self.query.results_iter(): File "/usr/lib/python2.6/site-packages/django/db/models/sql/query.py", line 206, in results_iter for rows in self.execute_sql(MULTI): File "/usr/lib/python2.6/site-packages/django/db/models/sql/query.py", line 1734, in execute_sql cursor.execute(sql, params) OperationalError: server closed the connection unexpectedly This probably means the server terminated abnormally before or while processing the request.

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• Apps not showing in Django admin site

  - by jack

  I have a Django project with about 10 apps in it. But the admin interface only shows Auth and Site models which are part of Django distribution. Yes, the admin interface is up and working but none of my self-written apps shows there. INSTALLED_APPS INSTALLED_APPS = ( 'django.contrib.auth', 'django.contrib.sites', 'django.contrib.contenttypes', 'django.contrib.humanize', 'django.contrib.sessions', 'django.contrib.admin', 'django.contrib.admindocs', 'project.app1', ... app1/admin.py from django.contrib import admin from project.app1.models import * admin.site.register(model1) admin.site.register(model2) admin.site.register(model3) What could be wrong in this case? Looks like everything is configured as what document says. Thank you in advance.

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• Django and App Engine

  - by notnoop

  I wanted to check the status of running Django on the Google App Engine currently and what the benefits of running django on GAE over simply using Webapp. Django main killer feature, IMHO, is the reuseable apps and middleware. Unfortunately, most current Django apps use models or model forms (django-tags, django-reviews, django-profiles, Pinax apps). So what are the remaining features or benefits that django has that can still run in Google App Engine (other than what's disabled: the popular django apps, session and authentication middleware, users and admin, models, etc). Also, is there a list of the Django apps that work in App Engine as well?

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• review on django book vs django tutorial

  - by momo

  going through both the django book and tutorial, am a bit confused to the differences in approach (aren't they both written by the same people?) can anyone who has experience in both give a short review on them? i have decent python skills (largely untested though), but no experience at all in web apps and am trying to decide which one to stick to. i briefly looked in to practical django projects but that was a bit too complicated for me, my background is primarily bash scripting, the python i know i learned from an instant hacking tutorial and diving into python.

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• Django: How do I position a page when using Django templates

  - by swisstony

  I have a web page where the user enters some data and then clicks a submit button. I process the data and then use the same Django template to display the original data, the submit button, and the results. When I am using the Django template to display results, I would like the page to be automatically scrolled down to the part of the page where the results begin. This allows the user to scroll back up the page if she wants to change her original data and click submit again. Hopefully, there's some simple way of doing this that I can't see at the moment.

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• Modify Django Forms

  - by Ninefingers

  Hi All, I've recently been developing on the django platform and have stumbled upon Django Forms (forms.Form/forms.ModelForm) as ways of creating <form> html. Now, this is brilliant for quick stuff but what I'm trying to do is a little bit more complicated. Consider a DateField - my current form has fields for day, month and year and constructs a python date object from that. However, a django form creates a single textbox in which the correct format (say 2010-06-15) must be entered. As another example, for large fields I need to replace <input> with <textarea>. I'd like to take advantage of Django's forms for simple validation but I need something simpler for my users. So my question is: can I intercept the rendering of one of these objects to write out the html as I like? If so, do I have to do all the writing myself or can I only do those objects I wish to re-write? Thanks in advance.

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• Django Form Preview

  - by Mark Kecko

  I'm trying to use django's FormPreview and I can't get it to work properly. Here's my code: forms.py class MyForm(forms.ModelForm): status = forms.TypedChoiceField( coerce=int, choices=LIST_STATUS, label="type", widget=forms.RadioSelect ) description = forms.CharField(widget = forms.Textarea) stage = forms.CharField() def __init__(self, useradd=None, *args, **kwargs): super(MyForm, self).__init__(*args, **kwargs) self.fields['firm'].label = "Firm" class Meta: model = MyModel fields = ['status', 'description', 'stage'] class MyFormPreview(FormPreview): form_template = 'templates/post.html' preview_template = 'templates/review.html' def process_preview(self, request, cleaned_data): print "processed" def done(self, request, cleaned_data): print "done" # Do something with the cleaned_data, then redirect # to a "success" page. return HttpResponseRedirect('/') urls.py (r'^post/$', MyFormPreview(MyForm)), post.html <form id = "post_ad" action = "" method = "POST" enctype="multipart/form-data"> <table> {{form.as_table}} </table> <input type="submit" name="save" value="Post" /> </form> When I go to /post/ I get the correct form and I fill it out. When I submit the form it goes right back to /post/ but but there are no errors (I've tried displaying {{errors}}) and the form is empty. None of my print statements execute. I'm not sure what I'm missing. Can anyone help me out? I can't find any documentation besides what's on the django site. Also, what's the "preview" variable called that I should use in my preview.html template? {{preview}} or do I just do {{form}} again? -- Answered below. I tried adding 'django.contrib.formtools' to my installed_apps in settings and I tried using the code from the default form templates from django.contrib as suggested below. Still, when I submit the form I go right back to the post template, none of my print statements execute :(

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• Error in {% markdown %} filter in Django Nonrel

  - by Robert Smith

  I'm having trouble using Markdown in Django Nonrel. I followed this instructions (added 'django.contrib.markup' to INSTALLED_APPS, include {% load markup %} in the template and use |markdown filter after installing python-markdown) but I get the following error: Error in {% markdown %} filter: The Python markdown library isn't installed. In this line: /path/to/project/django/contrib/markup/templatetags/markup.py in markdown they will be silently ignored. """ try: import markdown except ImportError: if settings.DEBUG: raise template.TemplateSyntaxError("Error in {% markdown %} filter: The Python markdown library isn't installed.") ... return force_unicode(value) else: # markdown.version was first added in 1.6b. The only version of markdown # to fully support extensions before 1.6b was the shortlived 1.6a. if hasattr(markdown, 'version'): extensions = [e for e in arg.split(",") if e] It seems obvious that import markdown is causing the problem but when I run: $ python manage.py shell >>> import elementtree >>> import markdown everthing works alright. Running Markdown 2.0.3, Django 1.3.1, Python 2.7. UPDATE: I thought maybe this was an issue related to permissions, so I changed my project via chmod 777 -R, but it didn't work. Ideas? Thanks!

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• Django Template tag, generating template block tag

  - by Issy

  Hi Guys, Currently a bit stuck, wondering if anyone can assist. I am using django-adminfiles. Which is a near little application. I want to use it to insert images into posts/articles/pages for a site i am building. How django-adminfiles works is it inserts a placeholder i.e <<< ImageFile and this gets rendered using a django template. It also has the feature of inserting custom options i.e (Insert Medium Image) , i figured i would used this to automatically resize images and include it in the post (similar to how WP does it). Django-adminfiles makes use of sorl.thumbnail app to generate thumbnails. So i have tried testing generating thumbnails: The current template that is used to render the inserted image is: {% spaceless %} <img src="{{ upload.upload.url }}" width="{{ upload.width }}" height="{{ upload.height }}" class="{{ options.class }}" class="{{ options.size }}" alt="{% if options.alt %}{{ options.alt }}{% else %}{{ upload.title }}{% endif %}" /> {% endspaceless %} I tried modifying this to: {% load thumbnail %} {% spaceless %} <img src="{% thumbnail upload.upload.url 200x50 %}" width="{{ upload.width }}" height="{{ upload.height }}" class="{{ options.class }}" class="{{ options.size }}" alt="{% if options.alt %}{{ options.alt }}{% else %}{{ upload.title }}{% endif %}" /> {% endspaceless %} I get the error: Exception Value: Caught an exception while rendering: Source file: '/media/uploads/DSC_0014.jpg' does not exist. I figured the thumbnail needs the absolute path so tried putting that in the template, and that works. i.e this works: {% thumbnail '/Users/me/media/uploads/DSC_0014.jpg' 200x50 %} So basically i need to generate the absolute path to the file give the relative path (to web root). You could do this by passing the MEDIA_ROOT setting to the template, but the reason i want to do a template tag is to programmatically set the image size.

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