Why doesn't this work?

[John Dungan]

I'm try to pass a choice from a posted form to this generic django
view. The idea is to select the form rather than create a separate
generic request for each type.

def add_foo(request):
foo_type= request.POST.get('foo_type')
logging.info("FOO TYPE: %s", foo_type)
return create_object(request, form_class=foo_type,
post_save_redirect=reverse('policies.views.policy_list'),
)


The problem is that is fails on the logging stmt with:

Exception Value: 'unicode' object has no attribute '_meta'

I know this is going to be something simple, so my apologies. Also I
know the next line is trouble so any suggestions on passing the
correct form_type to create_object would be appreciated.

A latte on me at the next meeting to whoever solves this headache.

[Devin Venable]

John,

Logging seems to handle both regular and unicode strings.  No exceptions here using Python 2.5.2.

>>> foo_type = u'This is a unicode string'
>>> print type(foo_type)
<type 'unicode'>
>>> import logging
>>> logging.info("foo type: %s", foo_type)

Are you sure the exception is on the logging line? 

Also, which version of Python and Django are you using?  Or is this AppEngine behavior?

Devin

[Steven Osborn]

Sorry seems I sent this reply to John and not the group.


---------- Forwarded message ----------
From: Steven Osborn <osborn...@gmail.com>
Date: Wed, Apr 15, 2009 at 11:27 PM
Subject: Re: [tulsapython] Why doesn't this work?
To: John Dungan <johna...@gmail.com>


From Django Docs: >>>

If you provide form_class, it should be a django.forms.ModelForm
subclass. Use this argument when you need to customize the model's
form. See the ModelForm docs for more information.

Otherwise, model should be a Django model class and the form used will
be a standard ModelForm for model.

<<<

objects of ModelForm or Model type will have a _meta attribute.

I would use an if - else to determine what type of object your creating like:

#start code
from myapp.models import Foo, Bar

if "foo" in request.POST.get('foo_type'):
  foo_type  = Foo;  # I don't think Django is expecting an instance
here but if so you'll want to do Foo()
else "bar" in request.POST.get('foo_type'):
 foo_type = Bar;

#endcode

You could use eval() to turn the string into an object but that is a
terrible idea if your app is ever going to be used by anyone besides
you.

Hope this helps.




--
Steven Osborn
http://steven.bitsetters.com

--
Steven Osborn
http://steven.bitsetters.com

[Steven Osborn]

And John replied to me. :-P

---------- Forwarded message ----------
From: gmail <johna...@gmail.com>
Date: Thu, Apr 16, 2009 at 4:21 AM
Subject: Re: [tulsapython] Why doesn't this work?

To: Steven Osborn <osborn...@gmail.com>


Works like a charm.

You were right about my plan to use a string, so I've skipped it.

Thanks for your help.

[John Dungan]

Steve's suggestion helped me move on, but I do need to point out that
for some reason the str and unicode would not work in the logging
stmt. I know it should, but it didn't. In the 'real' code I put a
simple log stmt right before this one and it worked and blew up on the
line in question. I have no doubt that the create_object would have
been a non-starter for the same reason, but it never got that far.