say you want to have a different list-template for app "blog" model "entries", just add /admin/blog/entries/change_list.html to your templates directory and customize it. we´re using that a lot to add buttons for specific models.
I didn´t figure out how to customize the view yet. I guess it´s not as flexible as I´d like it to be. the purpose of the admin interface, as I understand it, is that you ´re able to pre-fill the tables. for a more complex application (or a CMS), you may need to write your own views and templates (which is not so easy with the current custom manipulators ...). if you´re more specific with your question, it´s easier to answer ... patrick. Am 06.10.2006 um 22:18 schrieb Paul Barry: > > I'm still trying to understand the purpose of the admin interface. I > don't think STACKED does exactly what I want, but my question is more > of a broad one. Is it meant to be used with a complex application? > It seems to me that you can't make changes to the admin interface > because you have no control over the template or the view. Say I > wanted to create my own template and view for handling a specfic weird > object, is there even a way to add a link on the main admin page to my > custom view/template? I get the impression for some projects you have > to end up writing your own whole new custom admin interface, is that > true? > > On 10/3/06, Michael <[EMAIL PROTECTED]> wrote: >> >> >> On 10/3/06, Paul Barry <[EMAIL PROTECTED]> wrote: >>> >>> Is the admin interface customizable at all? Specifically: >>> >>> 1. Can you create a workflow, like step 1, step 2, step 3, then >>> save >> object? >>> >>> 2. Can you create a lookup select for a foreign key? For >>> example, if >>> I have an object that has a ForeignKey and there are 10,000 >>> instances >>> of the foreign key, I would like to pop-up a new window, have the >>> user >>> search for the foreign key instance, and then set the value back on >>> the original form, because: >>> >>> a) loading 10,000 options in a select will be slow >>> b) an admin user will most likely not be able to find the specific >>> foreign key instance they are looking for in a drop-down. This >>> can be >>> true with even smaller lists, where you might want to look for >>> instance based on a couple of properties, such as name or >>> date_published. >>> >> >> use "raw_id_admin=True," in your model, like : >> user=models.ForeignKey >> (User,edit_inline=models.STACKED,raw_id_admin=True,) >> >> >> -- >> -- >> Michael >> >>> >> > > > --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
