Opened 12 years ago
Last modified 12 years ago
#18463 closed Bug
Using len() in Paginator object can raise an error — at Initial Version
Reported by: | Owned by: | nobody | |
---|---|---|---|
Component: | Core (Other) | Version: | dev |
Severity: | Normal | Keywords: | |
Cc: | Triage Stage: | Accepted | |
Has patch: | no | Needs documentation: | no |
Needs tests: | no | Patch needs improvement: | no |
Easy pickings: | no | UI/UX: | no |
Description
I'm getting the following error if I try to run a piece of code through my views.py:
Traceback (most recent call last):
File "<console>", line 1, in <module>
File "/home/mp/webapps/django/rep/app/views.py", line 331, in search_species
print len(recs)
File "/home/mp/webapps/django/lib/python2.7/django/core/paginator.py", line 88, in len
return len(self.object_list)
File "/home/mp/webapps/django/lib/python2.7/django/db/models/query.py", line 87, in len
self._result_cache = list(self.iterator())
File "/home/mp/webapps/django/lib/python2.7/django/db/models/query.py", line 284, in iterator
model_cls = deferred_class_factory(self.model, skip)
File "/home/mp/webapps/django/lib/python2.7/django/db/models/query_utils.py", line 180, in deferred_class_factory
return type(name, (model,), overrides)
TypeError: type() argument 1 must be string, not unicode
Strangely, if I run a similar code directly in the Django shell, everything goes well. The code looks like this:
from app.models import MyClass qs = MyClass.objects.all() from django.core.paginator import Paginator paginator = Paginator(qs, 25) recs = paginator.page(1) print len(recs)
So you have to run this code through a method in views.py to get a crash in the last line.
I'm afraid I have no idea of what's going on and how to properly fix this, but if I change the offending line from the traceback by forcing str() in the name parameter it works:
return type(str(name), (model,), overrides)