The following worked for me:
model field:
type = models.CharField(max_length=10, radio_admin=True,
choices=CHOICE_LIST)
where CHOICE_LIST is a tuple of tuples, defined inside the model if
local, outside if you need it in multiple models, like:
CHOICE_LIST = (('dbasename1', 'displayname1'), ('dbasename2',
'displayname2'),)
radio_admin=True causes the list to be displayed as a radio select
list (instead of a select/drop list or something.) The docs at
http://www.djangoproject.com/documentation/model-api/ say the
radio_admin is new in the developmental version.
Hope this helps,
Michael
On Nov 20, 6:48 am, Christian Joergensen <[EMAIL PROTECTED]> wrote:
> David Marko wrote:
> > How to define choices for filed that works in admin interface? When I
> > define CharField as below, the list of choices is computed intially
> > and doesn't refresh when data changes in Choices table.
>
> > type = models.CharField(verbose_name='Type', maxlength=30, choices =
> > Choice.filter("xxxx"), null=False, blank=False)
>
> Normally, one would use a foreign key in this case:
>
> choice = models.ForeignKey(Choice)
>
> Regards,
>
> --
> Christian Joergensen | Linux, programming or web
> consultancyhttp://www.razor.dk | Visit us at:http://www.gmta.info
--~--~---------~--~----~------------~-------~--~----~
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
-~----------~----~----~----~------~----~------~--~---
