How to use a foreign key in Django? - django

Question

I have a video model in django. Currently when a video is saved, the current logged user is not saved in the model (Other fields have a value, except the user field). The form uses a generic create django view.
I would like to know how to save the current logged user when a new video is saved?
multimedia/models.py
from django.db import models
from django.contrib.auth.models import User
from django.forms import ModelForm
from django.core.exceptions import PermissionDenied
from django.db.models.signals import post_save
from django.contrib.auth.decorators import login_required
class Video(models.Model):
user = models.ForeignKey(User, related_name='+',blank=True, null=True)
title = models.CharField(max_length=200)
description = models.TextField()
created = models.DateTimeField('date created', auto_now_add=True, blank=False)
file_url = models.CharField(max_length=2000, blank=True)
file_name = models.CharField(max_length=255, blank=True)
file_uploaded = models.DateTimeField('date file uploaded', null=True, blank=True)
file_upload_speed = models.FloatField(null=True, blank=True)
file_size = models.IntegerField(null=True, blank=True)
def has_file(self):
return len(self.file_url) > 0
def __unicode__(self):
return self.title
#models.permalink
def get_absolute_url(self):
return ('multimedia_video_detail', [self.id])
class VideoForm(ModelForm):
class Meta:
model = Video
# fields = ('title', 'description')
exclude = ('file_url', 'file_name', 'file_uploaded','file_upload_speed', 'file_size')
multimedia/views.py
from django.conf.urls.defaults import patterns, include, url
from models import Video, VideoForm
# Uncomment the next two lines to enable the admin:
from django.contrib import admin
admin.autodiscover()
urlpatterns = patterns('',
url(r'^$','django.views.generic.list_detail.object_list',{ 'queryset': Video.objects.all() }, name='multimedia_video_list'),
url(r'^new$', 'django.views.generic.create_update.create_object',{ 'model': Video },name='multimedia_video_new'),
url(r'^(?P<object_id>[\d]*)$', 'django.views.generic.list_detail.object_detail',{ 'queryset': Video.objects.all() }, name='multimedia_video_detail'),
url(r'^(?P<object_id>[\d]*)/edit$','django.views.generic.create_update.update_object',{ 'form_class': VideoForm }, name='multimedia_video_edit'),
url(r'(?P<object_id>[\d]*)/delete$', 'django.views.generic.create_update.delete_object', { 'model': Video, 'post_delete_redirect': '/videos' }, name='multimedia_video_delete'),
url(r'^(?P<object_id>[\d]*)/upload$','multimedia.views.upload_video',name='multimedia_video_upload'),
url(r'^(?P<object_id>[\d]*)/upload/done$','multimedia.views.upload_video_done',name='multimedia_video_upload_done'),
)
Thanks for any help to solve this issue

Answer 1

Instead of directly using the generic creation view, you need to use a wrapper around it that pulls the current user out of the request. Additionally, you create a model form that accepts the user, and overrides the save() method to set the user value on the model instance.
def VideoForm(user):
class _wrapped(forms.ModelForm):
class Meta:
model = models.Video
def save(self, *args, **kwargs):
self.instance.user = user
super(_wrapped, self).save(*args, **kwargs)
return _wrapped
#login_required
def create(request):
return django.views.generic.create_update.create_object(form_class=VideoForm(request.user), name='multimedia_video_new')