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  • python/django problem with sessions and language

    - by freakish
    Hello everyone! I have the following problem: on the main page I can change language. New language is saved in request.session['django_language']. I also have SESSION_COOKIE_DOMAIN set to my site, so session should be inherited by subdomains. And it is, because after changing language I check request.session['django_language'] in subdomains and it's fine. Then I use django.middleware.locale.LocaleMiddleware to translate my pages. And it works perfectly... only on main site! If I change language and refresh main site - it is ok. However, if I change language and go to a subpage (for example /LogIn), then the page is NOT translated at all. It stays on default language. This is really strange, because if I use {% load i18n %} {% get_current_language as lang %} in this subpage, then lang is good language. There is no mistake. What kind of problem can it be? Some suggestions?

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  • How can I call model methods or properties from Django Admin?

    - by kg
    Is there a natural way to display model methods or properties in the Django admin site? In my case I have base statistics for a character that are part of the model, but other things such as status effects which affect the total calculation for that statistic: class Character(models.Model): base_dexterity = models.IntegerField(default=0) @property def dexterity(stat_name): total = self.base_dexterity total += sum(s.dexterity for s in self.status.all()]) return total It would be nice if I could display the total calculated statistic alongside the field to change the base statistic in the Change Character admin page, but it is not clear to me how to incorporate that information into the page.

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  • Django Multi-Table Inheritance VS Specifying Explicit OneToOne Relationship in Models

    - by chefsmart
    Hope all this makes sense :) I'll clarify via comments if necessary. Also, I am experimenting using bold text in this question, and will edit it out if I (or you) find it distracting. With that out of the way... Using django.contrib.auth gives us User and Group, among other useful things that I can't do without (like basic messaging). In my app I have several different types of users. A user can be of only one type. That would easily be handled by groups, with a little extra care. However, these different users are related to each other in hierarchies / relationships. Let's take a look at these users: - Principals - "top level" users Administrators - each administrator reports to a Principal Coordinators - each coordinator reports to an Administrator Apart from these there are other user types that are not directly related, but may get related later on. For example, "Company" is another type of user, and can have various "Products", and products may be supervised by a "Coordinator". "Buyer" is another kind of user that may buy products. Now all these users have various other attributes, some of which are common to all types of users and some of which are distinct only to one user type. For example, all types of users have to have an address. On the other hand, only the Principal user belongs to a "BranchOffice". Another point, which was stated above, is that a User can only ever be of one type. The app also needs to keep track of who created and/or modified Principals, Administrators, Coordinators, Companies, Products etc. (So that's two more links to the User model.) In this scenario, is it a good idea to use Django's multi-table inheritance as follows: - from django.contrib.auth.models import User class Principal(User): # # # branchoffice = models.ForeignKey(BranchOffice) landline = models.CharField(blank=True, max_length=20) mobile = models.CharField(blank=True, max_length=20) created_by = models.ForeignKey(User, editable=False, blank=True, related_name="principalcreator") modified_by = models.ForeignKey(User, editable=False, blank=True, related_name="principalmodifier") # # # Or should I go about doing it like this: - class Principal(models.Model): # # # user = models.OneToOneField(User, blank=True) branchoffice = models.ForeignKey(BranchOffice) landline = models.CharField(blank=True, max_length=20) mobile = models.CharField(blank=True, max_length=20) created_by = models.ForeignKey(User, editable=False, blank=True, related_name="principalcreator") modified_by = models.ForeignKey(User, editable=False, blank=True, related_name="principalmodifier") # # # Please keep in mind that there are other user types that are related via foreign keys, for example: - class Administrator(models.Model): # # # principal = models.ForeignKey(Principal, help_text="The supervising principal for this Administrator") user = models.OneToOneField(User, blank=True) province = models.ForeignKey( Province) landline = models.CharField(blank=True, max_length=20) mobile = models.CharField(blank=True, max_length=20) created_by = models.ForeignKey(User, editable=False, blank=True, related_name="administratorcreator") modified_by = models.ForeignKey(User, editable=False, blank=True, related_name="administratormodifier") I am aware that Django does use a one-to-one relationship for multi-table inheritance behind the scenes. I am just not qualified enough to decide which is a more sound approach.

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  • Django Forms - change the render multiple select widget

    - by John
    Hi, In my model I have a manytomany field mentors = models.ManyToManyField(MentorArea, verbose_name='Areas', blank=True) In my form I want to render this as: drop down box with list of all MentorArea objects which has not been associated with the object. Next to that an add button which will call a javascript function which will add it to the object. Then under that a ul list which has each selected MentorArea object with a x next to it which again calls a javascript function which will remove the MentorArea from the object. I know that to change how an field element is rendered you create a custom widget and override the render function and I have done that to create the add button. class AreaWidget(widgets.Select): def render(self, name, value, attrs=None, choices=()): jquery = u''' <input class="button def" type="button" value="Add" id="Add Area" />''' output = super(AreaWidget, self).render(name, value, attrs, choices) return output + mark_safe(jquery) However I don't know how to list the currently selected ones underneath as a list. Can anyone help me? Also what is the best way to filter down the list so that it only shows MentorArea objects which have not been added? I currently have the field as mentors = forms.ModelMultipleChoiceField(queryset=MentorArea.objects.all(), widget = AreaWidget, required=False) but this shows all mentors no matter if they have been added or not. Thanks

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  • django internationlisation

    - by ha22109
    Hello I need to have multiple language support of my django admin application.I can create the messege files.But how can i change the text of my models.The heading ,fields etc .I m only able to change the static elements which are there in my template. here is example of my class class Mymodel(model.Models): id=models.IntegerField(primary_key=true) name=models.CharField(max_length=200) group=models.CharField(max_length=200) class Meta: managed=False verbose_name_plural='My admin' db_table='my_admin' one more question.In my home page it is showing my verbose name 'My admin' which i mentioned.But when i go to list page it shows me the class name 'mymodel'.Why so.Can i changed that to

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  • Django: query spanning multiple many-to-many relationships

    - by Brant
    I've got some models set up like this: class AppGroup(models.Model): users = models.ManyToManyField(User) class Notification(models.Model): groups_to_notify = models.ManyToManyField(AppGroup) The User objects come from django's authentication system. Now, I am trying to get all the notifications pertaining to the groups that the current user is a part of. I have tried.. notifications = Notification.objects.filter(groups_to_notify=AppGroup.objects.filter(users=request.user)) But that gives an error: more than one row returned by a subquery used as an expression Which I suppose is because the groups_to_notify is checking against several groups. How can I grab all the notifications meant for the user based on the groups he is a part of?

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  • Designing a database for a user/points system? (in Django)

    - by AP257
    First of all, sorry if this isn't an appropriate question for StackOverflow. I've tried to make it as generalisable as possible. I want to create a database (MySQL, site running Django) that has users, who can be allocated a certain number of points for various types of action - it's a collaborative game. My requirements are to obtain: the number of points a user has the user's ranking compared to all other users and the overall leaderboard (i.e. all users ranked in order of points) This is what I have so far, in my Django models.py file: class SiteUser(models.Model): name = models.CharField(max_length=250 ) email = models.EmailField(max_length=250 ) date_added = models.DateTimeField(auto_now_add=True) def points_total(self): points_added = PointsAdded.objects.filter(user=self) points_total = 0 for point in points_added: points_total += point.points return points_total class PointsAdded(models.Model): user = models.ForeignKey('SiteUser') action = models.ForeignKey('Action') date_added = models.DateTimeField(auto_now_add=True) def points(self): points = Action.objects.filter(action=self.action) return points class Action(models.Model): points = models.IntegerField() action = models.CharField(max_length=36) However it's rapidly becoming clear to me that it's actually quite complex (in Django query terms at least) to figure out the user's ranking and return the leaderboard of users. At least, I'm finding it tough. Is there a more elegant way to do something like this? This question seems to suggest that I shouldn't even have a separate points table - what do people think? It feels more robust to have separate tables, but I don't have much experience of database design.

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  • How to do custom display and auto-select in django admin multi-select field?

    - by rsp
    I'm new to django, so please feel free to tell me if I'm doing this incorrectly. I am trying to create a django ordering system. My order model: class Order(models.Model): ordered_by = models.ForeignKey(User, limit_choices_to = {'groups__name': "Managers", 'is_active': 1}) in my admin ANY user can enter an order, but ordered_by must be someone in the group "managers" (this is the behavior I want). Now, if the logged in user happens to be a manager I want it to automatically fill in the field with that logged in user. I have accomplished this by: class OrderAdmin(admin.ModelAdmin): def formfield_for_foreignkey(self, db_field, request, **kwargs): if db_field.name == "ordered_by": if request.user in User.objects.filter(groups__name='Managers', is_active=1): kwargs["initial"] = request.user.id kwargs["empty_label"] = "-------------" return db_field.formfield(**kwargs) return super(OrderAdmin, self).formfield_for_foreignkey(db_field, request, **kwargs) This also works, but the admin puts the username as the display for the select box by default. It would be nice to have the user's real name listed. I was able to do it with this: class UserModelMultipleChoiceField(forms.ModelMultipleChoiceField): def label_from_instance(self, obj): return obj.first_name + " " + obj.last_name class OrderForm(forms.ModelForm): ordered_by = UserModelChoiceField(queryset=User.objects.all().filter(groups__name='Managers', is_active=1)) class OrderAdmin(admin.ModelAdmin): form = OrderForm My problem: I can't to both of these. If I put in the formfield_for_foreignkey function and add form = OrderForm to use my custom "UserModelChoiceField", it puts the nice name display but it won't select the currently logged in user. I'm new to this, but my guess is that when I use UserModelChoiceField it "erases" the info passed in via formfield_for_foreignkey. Do I need to use the super() function somehow to pass on this info? or something completely different?

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  • Best way to re-use the same django models and admin for multiple apps

    - by kepioo
    Given a reference app ( called guide), how can I create additional apps that will reuse the same model/admin/views than guide - the motivation behind is to be able to individually control each subapp. guide guideApp1 exact same models/admin/views than guide guideApp2 exact same models/admin/views than guide in the Admin site, I should have : 1 section for guideApp1 with all the tables defined in guide, that applies to guideApp1 1 section for guideApp12 with all the tables defined in guide, that applies to guideApp2

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  • Django version in GAE

    - by Alex
    I'm tring to use Django 1.1 in GAE, But when I uncomment use_library('django', '1.1') in this script import os os.environ['DJANGO_SETTINGS_MODULE'] = 'settings' from google.appengine.dist import use_library #use_library('django', '1.1') # Google App Engine imports. from google.appengine.ext.webapp import util # Force Django to reload its settings. from django.conf import settings settings._target = None import django.core.handlers.wsgi import django.core.signals import django.db import django.dispatch.dispatcher # Unregister the rollback event handler. django.dispatch.dispatcher.disconnect( django.db._rollback_on_exception, django.core.signals.got_request_exception) def main(): # Create a Django application for WSGI. application = django.core.handlers.wsgi.WSGIHandler() # Run the WSGI CGI handler with that application. util.run_wsgi_app(application) if __name__ == "__main__": main() I receives AttributeError: 'module' object has no attribute 'disconnect' What is going on?

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  • Is a many-to-many relationship with extra fields the right tool for my job?

    - by whichhand
    Previously had a go at asking a more specific version of this question, but had trouble articulating what my question was. On reflection that made me doubt if my chosen solution was correct for the problem, so this time I will explain the problem and ask if a) I am on the right track and b) if there is a way around my current brick wall. I am currently building a web interface to enable an existing database to be interrogated by (a small number of) users. Sticking with the analogy from the docs, I have models that look something like this: class Musician(models.Model): first_name = models.CharField(max_length=50) last_name = models.CharField(max_length=50) dob = models.DateField() class Album(models.Model): artist = models.ForeignKey(Musician) name = models.CharField(max_length=100) class Instrument(models.Model): artist = models.ForeignKey(Musician) name = models.CharField(max_length=100) Where I have one central table (Musician) and several tables of associated data that are related by either ForeignKey or OneToOneFields. Users interact with the database by creating filtering criteria to select a subset of Musicians based on data the data on the main or related tables. Likewise, the users can then select what piece of data is used to rank results that are presented to them. The results are then viewed initially as a 2 dimensional table with a single row per Musician with selected data fields (or aggregates) in each column. To give you some idea of scale, the database has ~5,000 Musicians with around 20 fields of related data. Up to here is fine and I have a working implementation. However, it is important that I have the ability for a given user to upload there own annotation data sets (more than one) and then filter and order on these in the same way they can with the existing data. The way I had tried to do this was to add the models: class UserDataSets(models.Model): user = models.ForeignKey(User) name = models.CharField(max_length=100) description = models.CharField(max_length=64) results = models.ManyToManyField(Musician, through='UserData') class UserData(models.Model): artist = models.ForeignKey(Musician) dataset = models.ForeignKey(UserDataSets) score = models.IntegerField() class Meta: unique_together = (("artist", "dataset"),) I have a simple upload mechanism enabling users to upload a data set file that consists of 1 to 1 relationship between a Musician and their "score". Within a given user dataset each artist will be unique, but different datasets are independent from each other and will often contain entries for the same musician. This worked fine for displaying the data, starting from a given artist I can do something like this: artist = Musician.objects.get(pk=1) dataset = UserDataSets.objects.get(pk=5) print artist.userdata_set.get(dataset=dataset.pk) However, this approach fell over when I came to implement the filtering and ordering of query set of musicians based on the data contained in a single user data set. For example, I could easily order the query set based on all of the data in the UserData table like this: artists = Musician.objects.all().order_by(userdata__score) But that does not help me order by the results of a given single user dataset. Likewise I need to be able to filter the query set based on the "scores" from different user data sets (eg find all musicians with a score 5 in dataset1 and < 2 in dataset2). Is there a way of doing this, or am I going about the whole thing wrong?

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  • Django User M2M relationship

    - by Antonio
    When trying to syncdb with the following models: class Contact(models.Model): user_from = models.ForeignKey(User,related_name='from_user') user_to = models.ForeignKey(User, related_name='to_user') class Meta: unique_together = (('user_from', 'user_to'),) User.add_to_class('following', models.ManyToManyField('self', through=Contact, related_name='followers', symmetrical=False)) I get the following error: Error: One or more models did not validate: auth.user: Accessor for m2m field 'following' clashes with related m2m field 'User.followers'. Add a related_name argument to the definition for 'following'. auth.user: Reverse query name for m2m field 'following' clashes with related m2m field 'User.followers'. Add a related_name argument to the definition for 'following'. auth.user: The model User has two manually-defined m2m relations through the model Contact, which is not permitted. Please consider using an extra field on your intermediary model instead. auth.user: Accessor for m2m field 'following' clashes with related m2m field 'User.followers'. Add a related_name argument to the definition for 'following'. auth.user: Reverse query name for m2m field 'following' clashes with related m2m field 'User.followers'. Add a related_name argument to the definition for 'following'.

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  • Django Admin: not seeing any app (permission problem?)

    - by Facundo
    I have a site with Django running some custom apps. I was not using the Django ORM, just the view and templates but now I need to store some info so I created some models in one app and enabled the Admin. The problem is when I log in the Admin it just says "You don't have permission to edit anything", not even the Auth app shows in the page. I'm using the same user created with syncdb as a superuser. In the same server I have another site that is using the Admin just fine. Using Django 1.1.0 with Apache/2.2.10 mod_python/3.3.1 Python/2.5.2, with psql (PostgreSQL) 8.1.11 all in Gentoo Linux 2.6.23 Any ideas where I can find a solution? Thanks a lot. UPDATE: It works from the development server. I bet this has something to do with some filesystem permission but I just can't find it. UPDATE2: vhost configuration file: <Location /> SetHandler python-program PythonHandler django.core.handlers.modpython SetEnv DJANGO_SETTINGS_MODULE gpx.settings PythonDebug On PythonPath "['/var/django'] + sys.path" </Location> UPDATE 3: more info /var/django/gpx/init.py exists and is empty I run python manage.py from /var/django/gpx directory The site is GPX, one of the apps is contable and lives in /var/django/gpx/contable the user apache is webdev group and all these directories and files belong to that group and have rw permission UPDATE 4: confirmed that the settings file is the same for apache and runserver (renamed it and both broke) UPDATE 5: /var/django/gpx/contable/init.py exists This is the relevan part of urls.py: urlpatterns = patterns('', (r'^admin/', include(admin.site.urls)), ) urlpatterns += patterns('gpx', (r'^$', 'menues.views.index'), (r'^adm/$', 'menues.views.admIndex'),

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  • Reordering fields in Django model

    - by Alex Lebedev
    I want to add few fields to every model in my django application. This time it's created_at, updated_at and notes. Duplicating code for every of 20+ models seems dumb. So, I decided to use abstract base class which would add these fields. The problem is that fields inherited from abstract base class come first in the field list in admin. Declaring field order for every ModelAdmin class is not an option, it's even more duplicate code than with manual field declaration. In my final solution, I modified model constructor to reorder fields in _meta before creating new instance: class MyModel(models.Model): # Service fields notes = my_fields.NotesField() created_at = models.DateTimeField(auto_now_add=True) updated_at = models.DateTimeField(auto_now=True) class Meta: abstract = True last_fields = ("notes", "created_at", "updated_at") def __init__(self, *args, **kwargs): new_order = [f.name for f in self._meta.fields] for field in self.last_fields: new_order.remove(field) new_order.append(field) self._meta._field_name_cache.sort(key=lambda x: new_order.index(x.name)) super(TwangooModel, self).__init__(*args, **kwargs) class ModelA(MyModel): field1 = models.CharField() field2 = models.CharField() #etc ... It works as intended, but I'm wondering, is there a better way to acheive my goal?

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  • Django 1.4 dependency when packaging a Precise application

    - by Caustic
    I am trying to package a program I wrote that depends on Django 1.4.1 in Ubuntu 12.04. As Django 1.4.1 isn't available in Precise I am wondering if it is best to: Package up Django 1.4.1 and drop it in my ppa OR write a script that wgets Django at build time and installs. OR Something better that I haven't thought of. I am still inexperienced with packaging and would appreciate some advice Thanks

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  • Django Model: many-to-many or many-to-one?

    - by knuckfubuck
    I'm just learning django and following a tutorial. I have a Link and a Bookmark. Unlike the tutorial I'm following, I would like a link to be associated with only one Bookmark, but a Bookmark can have multiple links. Is this the way to setup the model? class Link(models.Model): url = models.URLField(unique=True) bookmark = models.ForeignKey(Bookmark) class Bookmark(models.Model): title = models.CharField(maxlength=200) user = models.ForeignKey(User) links = models.ManyToManyField(Link)

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  • How can I display multiple django modelformset forms together?

    - by JT
    I have a problem with needing to provide multiple model backed forms on the same page. I understand how to do this with single forms, i.e. just create both the forms call them something different then use the appropriate names in the template. Now how exactly do you expand that solution to work with modelformsets? The wrinkle, of course, is that each 'form' must be rendered together in the appropriate fieldset. For example I want my template to produce something like this: <fieldset> <label for="id_base-0-desc">Home Base Description:</label> <input id="id_base-0-desc" type="text" name="base-0-desc" maxlength="100" /> <label for="id_likes-0-icecream">Want ice cream?</label> <input type="checkbox" name="likes-0-icecream" id="id_likes-0-icecream" /> </fieldset> <fieldset> <label for="id_base-1-desc">Home Base Description:</label> <input id="id_base-1-desc" type="text" name="base-1-desc" maxlength="100" /> <label for="id_likes-1-icecream">Want ice cream?</label> <input type="checkbox" name="likes-1-icecream" id="id_likes-1-icecream" /> </fieldset> I am using a loop like this to process the results for base_form, likes_form in map(None, base_forms, likes_forms): which works as I'd expect (I'm using map because the # of forms can be different). The problem is that I can't figure out a way to do the same thing with the templating engine. The system does work if I layout all the base models together then all the likes models after wards, but it doesn't meet the layout requirements.

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  • In Django Combobox choices, how do you lookup description from short value?

    - by MikeN
    In Django models/forms the choices for a combobox often look like this: food_choices = (("",""), ("1", "Falafel"), ("2", "Hummus"), ("3", "Eggplant Stuff, Babaganoush???"), So the value to be stored in the database will be 1/2/3, but the displayed value on the form will be the long description. When we are working in code outside a form, how can we quickly lookup the long description given the short value stored in the model? So I want to map short values to long values: print foo("1") "Falafel"

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  • Django App for Image heavy Magazine Publishing?

    - by stapler
    I'm about to begin work on a Django project for an image heavy non-profit "community arts" magazine. The magazine is published monthly with about 6-8 articles that include with 4-10 images. I've been looking around for other projects that people have started specifically for publishing in Django... Are there any publishing specific Django apps you'd recommend? At the moment the only thing I can find is django-newsroom which looks interesting. Currently I'm just Image Kit / Photologue to attach galleries to an Article Model... but I'd love to figure out a way to more fully integrate images into the article content. Thanks in advance

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  • Add data to Django form class using modelformset_factory

    - by dean
    I have a problem where I need to display a lot of forms for detail data for a hierarchical data set. I want to display some relational fields as labels for the forms and I'm struggling with a way to do this in a more robust way. Here is the code... class Category(models.Model): name = models.CharField(max_length=160) class Item(models.Model): category = models.ForeignKey('Category') name = models.CharField(max_length=160) weight = models.IntegerField(default=0) class Meta: ordering = ('category','weight','name') class BudgetValue(models.Model): value = models.IntegerField() plan = models.ForeignKey('Plan') item = models.ForeignKey('Item') I use the modelformset_factory to create a formset of budgetvalue forms for a particular plan. What I'd like is item name and category name for each BudgetValue. When I iterate through the forms each one will be labeled properly. class BudgetValueForm(forms.ModelForm): item = forms.ModelChoiceField(queryset=Item.objects.all(),widget=forms.HiddenInput()) plan = forms.ModelChoiceField(queryset=Plan.objects.all(),widget=forms.HiddenInput()) category = "" < assign dynamically on form creation > item = "" < assign dynamically on form creation > class Meta: model = BudgetValue fields = ('item','plan','value') What I started out with is just creating a dictionary of budgetvalue.item.category.name, budgetvalue.item.name, and the form for each budget value. This gets passed to the template and I render it as I intended. I'm assuming that the ordering of the forms in the formset and the querset used to genererate the formset keep the budgetvalues in the same order and the dictionary is created correctly. That is the budgetvalue.item.name is associated with the correct form. This scares me and I'm thinking there has to be a better way. Any help would be greatly appreciated.

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  • Get list of Unique many-to-many records from a queryset

    - by rsp
    My models: class Order(models.Model): ordered_by = models.ForeignKey(User) reasons = models.ManyToManyField(Reason) class Reason(models.Model): description = models.CharField() Basically a user creates an order and gives reasons for that order. ie, john has two orders (one for pencils because he is out AND because he doesn't like his current stock. a second order for pens because he is out). I want to print a list out of all reasons a user has placed his orders. So under john, it should print "he is out", "he doesn't like his current stock"; those two lines only. If I simply select all of john's orders, iterate through them and print out their "reasons" it'll print "he is out", "he doesn't like his current stock" and then "he is out" again. I don't want these duplicate values. How do I select a list of his reasons for ALL his orders so that the list has all unique rows?

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  • Django: ordering by backward related field property

    - by Silver Light
    Hello! I have two models related one-to-many: a Post and a Comment: class Post(models.Model): title = models.CharField(max_length=200); content = models.TextField(); class Comment(models.Model): post = models.ForeignKey('Post'); body = models.TextField(); date_added = models.DateTimeField(); I want to get a list of posts, ordered by the date of the latest comment. If I would write a custom SQL query it would look like this: SELECT `posts`.`*`, MAX(`comments`.`date_added`) AS `date_of_lat_comment` FROM `posts`, `comments` WHERE `posts`.`id` = `comments`.`post_id` GROUP BY `posts`.`id` ORDER BY `date_of_lat_comment` DESC How can I do same thing using django ORM?

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  • Edit the opposite side of a many to many relationship with django generic form

    - by Ed
    I have two models: class Actor(models.Model): name = models.CharField(max_length=30, unique = True) event = models.ManyToManyField(Event, blank=True, null=True) class Event(models.Model): name = models.CharField(max_length=30, unique = True) long_description = models.TextField(blank=True, null=True) In a previous question: http://stackoverflow.com/questions/2503243/django-form-linking-2-models-by-many-to-many-field, I created an EventForm with a save function: class EventForm(forms.ModelForm): class Meta: model = Event def save(self, commit=True): instance = forms.ModelForm.save(self) instance.actors_set.clear() for actor in self.cleaned_data['actors']: instance.actors_set.add(actors) return instance This allowed me to add m2m links from the other side of the defined m2m connection. Now I want to edit the entry. I've been using a generic function: def generic_edit(request, modelname, object_id): modelname = modelname.lower() form_class = form_dict[modelname] return update_object(request, form_class = form_class, object_id = object_id, template_name = 'createdit.html' ) but this pulls in all the info except the many-to-many selections saved to this object. I think I need to do something similar to this: http://stackoverflow.com/questions/1700202/editing-both-sides-of-m2m-in-admin-page, but I haven't figured it out. How do I use the generic update_object to edit the other side of many-to-many link?

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