Wiadomość napisana w dniu 2008-11-19, o godz. 15:15, przez Fabio Natali:
> I know I can use list_display to manage which fields will show up in > the table. The point is, I would like to add a column which is not a > field of my model, but a link, specifically a different link for each > row of my table, that is for each instance of my model. > > (I will use that link to print a pdf report of that specific instance > of my model. Do you think there are better or more canonical ways to > achieve this?) > > Any hints/urls on how to start are appreciated. http://docs.djangoproject.com/en/dev/ref/contrib/admin/#list-display Additionally, set allow_tags=True property of the method returning link. -- We read Knuth so you don't have to. - Tim Peters Jarek Zgoda, R&D, Redefine [EMAIL PROTECTED] --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
