I don't suppose a CommaSeparatedIntegerField would work, would it? You'd lose some querying options that you'd have with a foreign key relationship, but that may not be important to you.
On Jun 25, 10:37 am, Xan <[EMAIL PROTECTED]> wrote: > sorry models.list(integer, max=unlimited) > > On Jun 25, 5:37 pm, Xan <[EMAIL PROTECTED]> wrote: > > > Yes, I'm looking for that but without defining the ManyToManyField. > > Could I wrote something like: > > > > class A(models.Model): > > > integers = models.list(Integer, max=unlimited) > > > Thanks, > > > On Jun 23, 9:10 pm, "Richard Dahl" <[EMAIL PROTECTED]> wrote: > > > > I am a bit confused, are you looking for this: > > > > class IntegerRelation(models.Model): > > > content = models.IntegerField() > > > > class A(models.Model): > > > integers = models.ManyToManyField(IntegerRelation) > > > > -richard > > > > On 6/22/08, Xan <[EMAIL PROTECTED]> wrote: > > > > > Hi, > > > > > Suposing you want a model A with several and indetermined number of > > > > integer fields. Is there any possibility of defining A without > > > > creating a Integer class and then add ManyToManyField(Integer) in > > > > class A? > > > > > Thanks a lot! > > > > Xan. --~--~---------~--~----~------------~-------~--~----~ 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 [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~----------~----~----~----~------~----~------~--~---
