Closed GoogleCodeExporter closed 9 years ago
first off, grappelli 2.3+ requires django 1.3+ (resp. django trunk). I´m not
sure if this causes the issue, but I can´t reproduce it with a not-supported
django-version.
Original comment by sehmaschine
on 16 Mar 2011 at 12:29
I just updated to Django 1.3RC1
And still cannot create the block with a single model.
models = ('myapp.*',) produces a block with all of the models in myapp
models = ('myapp.mymodel',) does not show up in dashboard
models = ('myapp.mymodel.*',) also does not show up in dashboard (contrary to
my earlier experience)
Original comment by ndudenho...@gmail.com
on 16 Mar 2011 at 1:00
Original comment by sehmaschine
on 16 Mar 2011 at 8:06
the correct implementation is:
models = ('myapp.*',) produces a block with all of the models in myapp
models = ('myapp.models.*',) same as before
models = ('myapp.models.MyModel',) single model
does that answer your question? if yes, I´m going to update the docs.
Original comment by sehmaschine
on 17 Mar 2011 at 3:01
Yes, that works. Thank you for clarifying.
After looking back at the docs, they are not wrong. That was a foolish mistake
on my part (of course you need to specify 'models'), but some more explicit
clarification in the docs would be nice.
Original comment by ndudenho...@gmail.com
on 17 Mar 2011 at 3:09
This issue was closed by revision r1421.
Original comment by sehmaschine
on 17 Mar 2011 at 3:11
Original issue reported on code.google.com by
ndudenho...@gmail.com
on 16 Mar 2011 at 12:22