On Sun, Sep 6, 2009 at 6:20 PM, Daniel Roseman <dan...@roseman.org.uk>wrote:
> > On Sep 6, 8:39 am, Joshua Partogi <joshua.part...@gmail.com> wrote: > > Dear all, > > > > I have a model that has a foreign e.g > > > > class Foo(models.Model): > > bar = models.ForeignKey(Bar) > > > > class Bar(models.Model): > > name = models.CharField(max_length=50) > > > > This is my FooAdmin: > > class FooAdmin(admin.ModelAdmin): > > list_display = ['bar'] > > > > Now in the admin Foo's display list I want to have a link to go to bar > > display list. What would be the best way to do that in django? I tried to > > search in the doc but could not find a relevant howto on this issue. > > > > Thanks in advance for the insights. > > > > regards, > > > > Define a custom method on the FooAdmin class which returns the HTML of > a link to the Bar changelist with the relevant filter applied: > > def bar_link(self, obj): > return '<a href="/admin/myapp/bar/?foo=%s">See Bars</a>' % > obj.pk > bar_link.allow_tags = True > > See > http://docs.djangoproject.com/en/dev/ref/contrib/admin/#django.contrib.admin.ModelAdmin.list_display > Is there any built-in method in django where we can that url without having to hardcode it? -- http://blog.scrum8.com http://twitter.com/scrum8 --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~----------~----~----~----~------~----~------~--~---
