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• Overriding Django Admin's main page? - Django

  - by RadiantHex

  Hi folks, I'm trying to add features to Django admin's main page. I've been playing with index.html, but features added to this page affect all app pages. Any ideas on what template I'm supposed to use? Thanks loads!!

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• No Module named django.core

  - by Sirish Kumar

  Hi, I have updated to latest Django version 1.0.2 after uninstalling my old Django version.But now when I run django-admin.py I get the following error. How can I resolve this? Traceback (most recent call last): File "C:\Python25\Lib\site-packages\django\bin\django-admin.py", line 2, in <module> from django.core import management ImportError: No module named django.core

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• How to make a Django model fields calculated at runtime?

  - by Anatoly Rr

  I have a model: class Person (models.Model): name = models.CharField () birthday = models.DateField () age = models.IntegerField () I want to make age field to behave like a property: def get_age (self): return (datetime.datetime.now() - self.birthday).days // 365 age = property (get_age) but at the same time I need age to be a true field, so I can find it in Person._meta.fields, and assign attributes to it: age.help_text = "Age of the person", etc. Obviously I cannot just override Person.save() method to calculate and store age in the database, because it inevitably will become wrong later (in fact, it shouldn't be stored in the database at all). Actually, I don't need to have setters now, but a nice solution must have setting feature. Is it possible in Django, or probably there is a more pythonic and djangoic approach to my problem?

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• How to preprocess a Django model field value before return?

  - by Satoru.Logic

  Hi, all. I have a Note model class like this: class Note(models.Model): author = models.ForeignKey(User, related_name='notes') content = NoteContentField(max_length=256) NoteContentField is a custom sub-class of CharField that override the to_python method in purpose of doing some twitter-text-conversion processing. class NoteContentField(models.CharField): __metaclass__ = models.SubfieldBase def to_python(self, value): value = super(NoteContentField, self).to_python(value) from ..utils import linkify return mark_safe(linkify(value)) However, this doesn't work. When I save a Note object like this: note = Note(author=request.use, content=form.cleaned_data['content']) The conversed value is saved into the database, which is not what I wanna see. Would you please tell me what's wrong with this? Thanks in advance.

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• Pydev and Django: Shell not finding certain modules?

  - by Rosarch

  I am developing a Django project with PyDev in Eclipse. For a while, PyDev's Django Shell worked great. Now, it doesn't: >>> import sys; print('%s %s' % (sys.executable or sys.platform, sys.version)) C:\Python26\python.exe 2.6.4 (r264:75708, Oct 26 2009, 08:23:19) [MSC v.1500 32 bit (Intel)] >>> >>> from django.core import management;import mysite.settings as settings;management.setup_environ(settings) Traceback (most recent call last): File "<console>", line 1, in <module> ImportError: No module named mysite.settings >>> The dev server runs just fine. What could I be doing wrong? The models module is also conspicuously absent: >>> import mysite.myapp.models Traceback (most recent call last): File "<console>", line 1, in <module> ImportError: No module named mysite.myapp.models On the normal command line, outside of PyDev, the shell works fine. Why could this be happening?

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• Using django and django-voting app, how can I order a queryset according to the votes of each item?

  - by snz3

  (I'm new to python and django so please bear with me for a second. I apologise if this has been answered elsewhere and couldn't find it) Let's say I have a Link model and through the django-voting application users can vote on link instances. How can I order those link instances according to their score, eg. display those with the higher score first. I assume I could use the get_top manager of django-voting, but that would only give me the top scoring link instances and wouldn't take into consideration other parameters I would like to add (for example, those links that belong to a specific user or paging or whatever). My guess would be to write a custom manager for my Link model where by I can filter a queryset according to each item's score. If I understand correctly that will require me to loop through each item, check its score, and then place it a list (or dictionary) which will then be sorted according to the score of each item. That wouldn't return a queryset but a dictionary with each item. Am I missing something here?

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• django-social-auth for Facebook is redirecting home and not logging in

  - by Scott Rogowski

  I have had django-social-auth working for Google for quite some time now but am having problems with Facebook. I am at the point where clicking on the /login/facebook/ link will take me to the Facebook authorization page. I then click "go to app" and it redirects me to my home page but does not log in or create a user but does put some strange "#=" onto the back of my URL. Reading up on that, here https://developers.facebook.com/blog/post/552/, and here https://github.com/omab/django-social-auth/issues/199, it seems that would be happening if the redirect uri was not defined. However, on my facebook app settings, I have the following (replacing my site with example.com): + App Namespace: "example" + Site URL: "http://example.com/complete/facebook/" + Site Domain: "example.com" + Sandbox Mode: "On" + Post-Authorize Redirect URL: "http://apps.facebook.com/example/" + Deauthorize URL: "http://www.example.com/" + Post-Authorize URL: "http://example.com/complete/facebook/" The request that django-social-auth is sending to facebook is (replacing my info again): "https://www.facebook.com/dialog/oauth?scope=email&state=*&redirect_uri=http%3A%2F%2Fexample.com%2Fcomplete%2Ffacebook%2F%3Fredirect_state%3D***&client_id=*" The /complete/facebook/ is what is in the documentation and google works as /complete/google/ What am I missing here?

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• Django m2m form appearing fields

  - by dana

  I have a classroom application,and a follow relation. Users can follow each other and can create classrooms.When a user creates a classroom, he can invite only the people that are following him. The Classroom model is a m2m to User table. i have in models. py: class Classroom(models.Model): creator = models.ForeignKey(User) classname = models.CharField(max_length=140, unique = True) date = models.DateTimeField(auto_now=True) open_class = models.BooleanField(default=True) members = models.ManyToManyField(User,related_name="list of invited members") and in models.py of the follow application: class Relations(models.Model): initiated_by = models.ForeignKey(User, editable=False) date_initiated = models.DateTimeField(auto_now=True, editable = False) follow = models.ForeignKey(User, editable = False, related_name = "follow") date_follow = models.DateTimeField(auto_now=True, editable = False) and in views.py of the classroom app: def save_classroom(request, username): if request.method == 'POST': u = User.objects.get(username=username) form = ClassroomForm(request.POST, request.FILES) if form.is_valid(): new_obj = form.save(commit=False) new_obj.creator = request.user r = Relations.objects.filter(initiated_by = request.user) # new_obj.members = new_obj.save() return HttpResponseRedirect('.') else: form = ClassroomForm() return render_to_response('classroom/classroom_form.html', { 'form': form, }, context_instance=RequestContext(request)) i'm using a ModelForm for the classroom form, and the default view, taking in consideration my many to many relation with User table, in the field Members, is a list of all Users in my database. But i only want in that list the users that are in a follow relationship with the logged in user - the one who creates the classroom. How can i do that? Thanks!

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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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• Global name not defined error in Django/Python trying to set foreignkey

  - by Mark

  Summary: I define a method createPage within a file called PageTree.py that takes a Source model object and a string. The method tries to generate a Page model object. It tries to set the Page model object's foreignkey to refer to the Source model object which was passed in. This throws a NameError exception! I'm trying to represent a website which is structured like a tree. I define the Django models Page and Source, Page representing a node on the tree and Source representing the contents of the page. (You can probably skip over these, this is a basic tree implementation using doubly linked nodes). class Page(models.Model): name = models.CharField(max_length=50) parent = models.ForeignKey("self", related_name="children", null=True); firstChild = models.ForeignKey("self", related_name="origin", null=True); nextSibling = models.ForeignKey("self", related_name="prevSibling", null=True); previousSibling = models.ForeignKey("self", related_name="nxtSibling", null=True); source = models.ForeignKey("Source"); class Source(models.Model): #A source that is non dynamic will be refered to as a static source #Dynamic sources contain locations that are names of functions #Static sources contain locations that are places on disk name = models.CharField(primary_key=True, max_length=50) isDynamic = models.BooleanField() location = models.CharField(max_length=100); I've coded a python program called PageTree.py which allows me to request nodes from the database and manipulate the structure of the tree. Here is the trouble making method: def createPage(pageSource, pageName): page = Page() page.source = pageSource page.name = pageName page.save() return page I'm running this program in a shell through manage.py in Windows 7 manage.py shell from mysite.PageManager.models import Page, Source from mysite.PageManager.PageTree import * ... create someSource = Source(), populate the fields, and save it ... createPage(someSource, "test") ... NameError: global name 'source' is not defined When I type in the function definition for createPage into the shell by hand, the call works without error. This is driving me bonkers and help is appreciated.

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• Django admin, hide a model

  - by Hellnar

  Hello At the root page of the admin site where registered models appear, I want to hide several models that are registered to the Django admin. If I directly unregister those, I am not able to add new records as the add new symbol "+" dissapears. How can this be done ?

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• working on lists in python

  - by owca

  I'm trying to make a small modification to django lfs project, that will allow me to deactivate products with no stocks. Unfortunatelly I'm just beginning to learn python, so I have big trouble with its syntax. That's what I'm trying to do. I'm using method 'has_variants' which returns true if product has any. Then I'm building a list from variants for this product. Next for every product in this list (I've called it 'set') I check it's stock and set bool variable 'inactive' to true if product has no stocks and to false if there are any. Finally if 'inactive' is false I'm setting self.active to 0. Code fails in line with: set[] = s How to correct it ? def deactivate(self): """If there are no stocks, deactivate the product. Used in last step of checkout. """ if self.has_variants(): for s in self.variants.filter(active=True): set[] = s for var in set: if var.get_stock_amount() == 0: inactive = True else: inactive = False else: if self.get_stock_amount() == 0: inactive = True if inactive: self.active = False return 0 error log : Traceback (most recent call last): File "manage.py", line 11, in <module> execute_manager(settings) File "/home/purplecow/rails/purpledev/site-packages/django/core/management/__i nit__.py", line 362, in execute_manager utility.execute() File "/home/purplecow/rails/purpledev/site-packages/django/core/management/__i nit__.py", line 303, in execute self.fetch_command(subcommand).run_from_argv(self.argv) File "/home/purplecow/rails/purpledev/site-packages/django/core/management/bas e.py", line 195, in run_from_argv self.execute(*args, **options.__dict__) File "/home/purplecow/rails/purpledev/site-packages/django/core/management/bas e.py", line 213, in execute translation.activate('en-us') File "/home/purplecow/rails/purpledev/site-packages/django/utils/translation/_ _init__.py", line 73, in activate return real_activate(language) File "/home/purplecow/rails/purpledev/site-packages/django/utils/translation/_ _init__.py", line 43, in delayed_loader return g['real_%s' % caller](*args, **kwargs) File "/home/purplecow/rails/purpledev/site-packages/django/utils/translation/t rans_real.py", line 205, in activate _active[currentThread()] = translation(language) File "/home/purplecow/rails/purpledev/site-packages/django/utils/translation/t rans_real.py", line 194, in translation default_translation = _fetch(settings.LANGUAGE_CODE) File "/home/purplecow/rails/purpledev/site-packages/django/utils/translation/t rans_real.py", line 180, in _fetch app = import_module(appname) File "/home/purplecow/rails/purpledev/site-packages/django/utils/importlib.py" , line 35, in import_module __import__(name) File "/home/purplecow/rails/purpledev/lfs/caching/__init__.py", line 1, in <mo dule> from listeners import * File "/home/purplecow/rails/purpledev/lfs/caching/listeners.py", line 10, in < module> from lfs.cart.models import Cart File "/home/purplecow/rails/purpledev/lfs/cart/models.py", line 8, in <module> from lfs.catalog.models import Product File "/home/purplecow/rails/purpledev/lfs/catalog/__init__.py", line 1, in <mo dule> from listeners import * File "/home/purplecow/rails/purpledev/lfs/catalog/listeners.py", line 5, in <m odule> from lfs.catalog.models import PropertyGroup File "/home/purplecow/rails/purpledev/lfs/catalog/models.py", line 589 set[] = s ^ SyntaxError: invalid syntax

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• Django admin's filter_horizontal (& filter_vertical) not working

  - by negus

  I'm trying to use ModelAdmin.filter_horizontal and ModelAdmin.filter_vertical for ManyToMany field instead of select multiple box but all I get is: My model: class Title(models.Model): #... production_companies = models.ManyToManyField(Company, verbose_name="????????-?????????????") #... My admin: class TitleAdmin(admin.ModelAdmin): prepopulated_fields = {"slug": ("original_name",)} filter_horizontal = ("production_companies",) radio_fields = {"state": admin.HORIZONTAL} #... The javascripts are loading OK, I really don't get what happens. Django 1.1.1 stable.

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• SelectDateWidget in Django Admin?

  - by Maria

  Can i change default AdminDateWidget to SelectDateWidget in my models? How can i do this? I try: class RespondentAdmin(admin.ModelAdmin): formfield_overrides = { models.DateField: {'widget': SelectDateWidget}, } but it doesn't work

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• Django: Template should render 'description' not actual value

  - by Till Backhaus

  Hi, in a Model I have a CharField with choices: class MyModel(models.Model): THE_CHOICES=( ('val',_(u'Value Description')), ) ... myfield=models.CharField(max_length=3,choices=THE_CHOICES Now in the template I access an instance of MyModel: {{ my_instance.myfield }} Of course the gives me val instead of Value Description. How do I get the description? Thanks in advance!

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• Django Custom Admin

  - by Harry

  Hello there Currently I have an app with a save method in its models.py . So I would like to change this so the models does not do the save() method, but rather have a separate view that will do this in die admin site. Can you please direct me in the correct direction? Thank you

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• Django modelform ForeignKey List

  - by Harry

  How do you get each item in the ForeignKey field in a list, for example: class Delegate(models.Model): excursion = models.ForeignKey(Excursion, limit_choices_to = {'is_activity': False}, related_name='excursion', null=True, blank=True) Template: {% for object in formset.excursion_set.all %} {{ object.lable }} etc {% endfor %} My reason is that I don't want the options to display as a dropdown, but in a custom way that I will style etc.

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• Implementing a popularity algorithm in Django

  - by TheLizardKing

  I am creating a site similar to reddit and hacker news that has a database of links and votes. I am implementing hacker news' popularity algorithm and things are going pretty swimmingly until it comes to actually gathering up these links and displaying them. The algorithm is simple: Y Combinator's Hacker News: Popularity = (p - 1) / (t + 2)^1.5` Votes divided by age factor. Where` p : votes (points) from users. t : time since submission in hours. p is subtracted by 1 to negate submitter's vote. Age factor is (time since submission in hours plus two) to the power of 1.5.factor is (time since submission in hours plus two) to the power of 1.5. I asked a very similar question over yonder http://stackoverflow.com/questions/1964395/complex-ordering-in-django but instead of contemplating my options I choose one and tried to make it work because that's how I did it with PHP/MySQL but I now know Django does things a lot differently. My models look something (exactly) like this class Link(models.Model): category = models.ForeignKey(Category) user = models.ForeignKey(User) created = models.DateTimeField(auto_now_add = True) modified = models.DateTimeField(auto_now = True) fame = models.PositiveIntegerField(default = 1) title = models.CharField(max_length = 256) url = models.URLField(max_length = 2048) def __unicode__(self): return self.title class Vote(models.Model): link = models.ForeignKey(Link) user = models.ForeignKey(User) created = models.DateTimeField(auto_now_add = True) modified = models.DateTimeField(auto_now = True) karma_delta = models.SmallIntegerField() def __unicode__(self): return str(self.karma_delta) and my view: def index(request): popular_links = Link.objects.select_related().annotate(karma_total = Sum('vote__karma_delta')) return render_to_response('links/index.html', {'links': popular_links}) Now from my previous question, I am trying to implement the algorithm using the sorting function. An answer from that question seems to think I should put the algorithm in the select and sort then. I am going to paginate these results so I don't think I can do the sorting in python without grabbing everything. Any suggestions on how I could efficiently do this? EDIT This isn't working yet but I think it's a step in the right direction: from django.shortcuts import render_to_response from linkett.apps.links.models import * def index(request): popular_links = Link.objects.select_related() popular_links = popular_links.extra( select = { 'karma_total': 'SUM(vote.karma_delta)', 'popularity': '(karma_total - 1) / POW(2, 1.5)', }, order_by = ['-popularity'] ) return render_to_response('links/index.html', {'links': popular_links}) This errors out into: Caught an exception while rendering: column "karma_total" does not exist LINE 1: SELECT ((karma_total - 1) / POW(2, 1.5)) AS "popularity", (S... EDIT 2 Better error? TemplateSyntaxError: Caught an exception while rendering: missing FROM-clause entry for table "vote" LINE 1: SELECT ((vote.karma_total - 1) / POW(2, 1.5)) AS "popularity... My index.html is simply: {% block content %} {% for link in links %} karma-up {{ link.karma_total }} karma-down {{ link.title }} Posted by {{ link.user }} to {{ link.category }} at {{ link.created }} {% empty %} No Links {% endfor %} {% endblock content %} EDIT 3 So very close! Again, all these answers are great but I am concentrating on a particular one because I feel it works best for my situation. from django.db.models import Sum from django.shortcuts import render_to_response from linkett.apps.links.models import * def index(request): popular_links = Link.objects.select_related().extra( select = { 'popularity': '(SUM(links_vote.karma_delta) - 1) / POW(2, 1.5)', }, tables = ['links_link', 'links_vote'], order_by = ['-popularity'], ) return render_to_response('links/test.html', {'links': popular_links}) Running this I am presented with an error hating on my lack of group by values. Specifically: TemplateSyntaxError at / Caught an exception while rendering: column "links_link.id" must appear in the GROUP BY clause or be used in an aggregate function LINE 1: ...karma_delta) - 1) / POW(2, 1.5)) AS "popularity", "links_lin... Not sure why my links_link.id wouldn't be in my group by but I am not sure how to alter my group by, django usually does that.

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• Django Per-site caching using memcached

  - by Paul

  Hi, So I'm using per-site caching on a project and I've observed the following, which is kind of confusing. When I load a flat page in my browser then change it through admin and then do a refresh (within the cache timeout) there is no change in the page--as expected. However when I stat a new session in a different browser and load the page (still within the timeout) the app is hit instead of the cache, with the Isn't the cache key generated from the URL? it seems that the session state is getting in there somewhere, which is causing a cache miss. Any ideas? thanks MIDDLEWARE_CLASSES = ( 'django.middleware.cache.UpdateCacheMiddleware', 'django.middleware.common.CommonMiddleware', 'django.contrib.sessions.middleware.SessionMiddleware', 'django.contrib.auth.middleware.AuthenticationMiddleware', 'django.middleware.gzip.GZipMiddleware', 'django.middleware.http.ConditionalGetMiddleware', 'django.middleware.doc.XViewMiddleware', 'ittybitty.middleware.IttyBittyURLMiddleware', 'django.contrib.flatpages.middleware.FlatpageFallbackMiddleware', 'maintenancemode.middleware.MaintenanceModeMiddleware', 'djangodblog.middleware.DBLogMiddleware', 'SSL.middleware.SSLRedirect', #SSL middleware to handle SSL 'django.middleware.cache.FetchFromCacheMiddleware', )

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• I keep Getting KeyError: 'tried' Whenever I Tried to Run Django Dev Server from Remote Machine

  - by Spikie

  I am running django on python2.6.1, and did start the django web server like this manage.py runserver 192.0.0.1:8000 then tried to connect to the django dev web server on http://192.0.0.1:8000/ keep getting this message on the remote computer Traceback (most recent call last): File "C:\Python26\Lib\site-packages\django\core\servers\basehttp.py", line 279, in run self.result = application(self.environ, self.start_response) File "C:\Python26\Lib\site-packages\django\core\servers\basehttp.py", line 651, in call return self.application(environ, start_response) File "C:\Python26\lib\site-packages\django\core\handlers\wsgi.py", line 241, in call response = self.get_response(request) File "C:\Python26\lib\site-packages\django\core\handlers\base.py", line 115, in get_response return debug.technical_404_response(request, e) File "C:\Python26\Lib\site-packages\django\views\debug.py", line 247, in technical_404_response tried = exception.args[0]['tried'] KeyError: 'tried' what i am doing wrong ? it seen to work ok if i run http://192.0.0.1:8000/ on the computer that runs the Django web server and have that ip 192.0.0.1:8000

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• Testing a Django view cause "AttributeError: 'NoneType' object has no attribute 'handler500'" error

  - by jack

  I just wanted to start testing a Django view using the code below: from django.test.client import Client c = Client() response = c.get('/search/keyword') print response.content It just throws out following error message: "/usr/local/lib/python2.6/dist-packages/django/test/client.py", line 286, in get response = self.request(**r) File "/usr/local/lib/python2.6/dist-packages/django/test/client.py", line 230, in request response = self.handler(environ) File "/usr/local/lib/python2.6/dist-packages/django/test/client.py", line 74, in __call__ response = self.get_response(request) File "/usr/local/lib/python2.6/dist-packages/django/core/handlers/base.py", line 143, in get_response return self.handle_uncaught_exception(request, resolver, exc_info) File "/usr/local/lib/python2.6/dist-packages/django/core/handlers/base.py", line 178, in handle_uncaught_exception callback, param_dict = resolver.resolve500() File "/usr/local/lib/python2.6/dist-packages/django/core/urlresolvers.py", line 268, in resolve500 return self._resolve_special('500') File "/usr/local/lib/python2.6/dist-packages/django/core/urlresolvers.py", line 258, in _resolve_special callback = getattr(self.urlconf_module, 'handler%s' % view_type) AttributeError: 'NoneType' object has no attribute 'handler500' The view works in browser. What's wrong with above code?

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• Ditching Django's models for Ajax/Web Services

  - by Igor Ganapolsky

  Recently I came across a problem at work that made me rethink Django's models. The app I am developing resides on a Linux server. Its a simple model/view/controller app that involves user interaction and updating data in the database. The problem is that this data resides in a MS SQL database on a Windows machine. So in order to use Django's models, I would have to leverage an ODBC driver on linux, and the use a python add-on like pyodbc. Well, let me tell you, setting up a reliable and functional ODBC connection on linux is no easy feat! So much so, that I spent several hours maneuvering this on my CentOS with no luck, and was left with frustration and lots of dumb system errors. In the meantime I have a deadline to meet, and suddenly the very agile and rapid Django application is a roadblock rather than a pleasure to work with. Someone on my team suggested writing this app in .NET. But there are a few problems with that: it won't be deployable on a linux machine, and I won't be able to work on it since I don't know ASP.net. Then a much better suggestion was made: keep the app in django, but instead of using models, do straight up ajax/web services calls in the template. And then it dawned on me - what a great idea. Django's models seem like a nuissance and hindrance in this case, and I can just have someone else write .Net services on their side, that I can call from my template. As a result my app will be leaner and more compact. So, I was wondering if you guys ever came across a similar dillema and what you decided to do about it.

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• Validation on ManyToManyField before Save in Models.py

  - by Heyl1

  I have the following models: class Application(models.Model): users = models.ManyToManyField(User, through='Permission') folder = models.ForeignKey(Folder) class Folder(models.Model): company = models.ManyToManyField(Compnay) class UserProfile(models.Model): user = models.OneToOneField(User, related_name='profile') company = models.ManyToManyField(Company) What I would like to do is to check whether one of the users of the Application has the same company as the Application (via Folder). If this is the case the Application instance should not be saved. The problem is that the ManyToManyFields aren't updated until after the 'post-save' signal. The only option seems to be the new m2m_changed signal. But I'm not sure how I then roll back the save that has already happened. Another option would be to rewrite the save function (in models.py, because I'm talking about the admin here), but I'm not sure how I could access the manytomanyfield content. Finally I've read something about rewriting the save function in the admin of the model in admin.py, however I still wouldn't know how you would access the manytomanyfield content. I have been searching for this everywhere but nothing I come across seems to work for me. If anything is unclear, please tell me. Thanks for your help! Heleen

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• django threadedcomments

  - by Patrick

  Hi folks, I would like to setup a comment systems on my site, using django threadedcomments, and I follow all the steps in the Tutorial, however, I get the following error: No module named newforms.util I am not sure what causing this issue, here is my configuration: #settings.py INSTALLED_APPS = ( 'django.contrib.admin', 'django.contrib.auth', 'django.contrib.contenttypes', 'django.contrib.sessions', 'django.contrib.sites', 'myproject.myapp', 'threadedcomments', ) #urls.py from django.conf import settings from django.conf.urls.defaults import * from django.contrib import admin admin.autodiscover() urlpatterns = patterns('', (r'^admin/', include(admin.site.urls)), (r'^threadedcomments/', include('threadedcomments.urls')), ) Please let me know if there is another better choice for commenting, as long as the comment system is flexible and able to do lot of customization, as well as threadedcomment, of coz, integrating with Rating, I am happy to use the other one. Thanks guys.

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• How do I reference the other object in django models

  - by UserZero

  Hi, first post here.In Django, I want to have many files be associated with a particular model, so I'm doing a seperate model called files and have model 'A' 'have many' files. But I want my files to be saved in director named by model 'A'. So For example I want something like this: class Show(models.Model): name = models.CharField() showfolder = models.FilePathField() class Episode(models.Model): show = models.ForeignKey(Show) name = models.CharField() files = models.ManyToManyField(mp3) class Mp3(models.Model): file = FileField(upload_to=Episode.show.showfolder) So hopefully that last line expresses what I WANT it to do(get the folder name from Show object associated with the episode). The question is how would I really write that?(besides jumping through hoops in the controller.) Thanks.

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