Hi,
I had a problem with the admin interface and found a related issue
from a trac ticket. I sent my below email to the users' list but
didn't get any reply. I guess you guys here have more insight of
Django and maybe can point me to the right direction? I have been
struggling with this for 2 days now.
I'm not sure if it's appropriate to post this message here. Do
apologize if it's not.
Many thanks
Cliff
On Jun 7, 11:41 pm, Cliff <[EMAIL PROTECTED]> wrote:
> Just to throw in another variable. The recursion comes from file
> django/db/models.py in method get_follow:
> def get_follow(self, override=None):
> follow = {}
> for f in self.fields + self.many_to_many +
> self.get_all_related_objects():
> if override and f.name in override:
> child_override = override[f.name]
> else:
> child_override = None
> fol = f.get_follow(child_override)
> if fol != None:
> follow[f.name] = fol
> return follow
> The same issue was reported in this
> tickethttp://code.djangoproject.com/ticket/3288
> and said to be fixed in this
> changesethttp://code.djangoproject.com/changeset/7477#file19
>
> Please if someone can throw some insight? Really appreciate your help.
> Cliff
>
> On Jun 7, 6:28 pm, Cliff <[EMAIL PROTECTED]> wrote:
>
> > Hi,
> > This is the simplified version of my code:
> > class Test(models.Model):
> > parent = models.ForeignKey('self',edit_inline=models.TABULAR)
>
> > and the error I got:
> > RuntimeError: maximum recursion depth exceeded
>
> > I can understand that's because it's doing recursive for the child
> > object, grand child object and so on until it reaches the upper limit
> > of recursion. What I really want to do is just one level of recursion,
> > so I get a structure of the following if know what I mean:
> > test(parent)
> > --test1(child)
> > --test2(child)
> > --test3(child)
>
> > Is there a way to achieve this?
> > Thanks
> > Cliff
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