I have an error message on django 1.4:
dictionary update sequence element #0 has length 1; 2 is required
[EDIT]
It happe
I got the same issue and found that it was due to wrong parameters.
In views.py
, I used:
return render(request, 'demo.html',{'items', items})
But I found the issue: {'items', items}
. Changing to {'items': items}
resolved the issue.
Just ran into this problem. I don't know if it's the same thing that hit your code, but for me the root cause was because I forgot to put name=
on the last argument of the url
(or path
in Django 2.0+) function call.
For instance, the following functions throw the error from the question:
url(r'^foo/(?P<bar>[A-Za-z]+)/$', views.FooBar.as_view(), 'foo')
path('foo/{slug:bar}/', views.FooBar, 'foo')
But these actually work:
url(r'^foo/(?P<bar>[A-Za-z]+)/$', views.FooBar.as_view(), name='foo')
path('foo/{slug:bar}/', views.FooBar, name='foo')
The reason why the traceback is unhelpful is because internally, Django wants to parse the given positional argument as the keyword argument kwargs
, and since a string is an iterable, an atypical code path begins to unfold. Always use name=
on your urls!
I hit this error calling:
dict(my_data)
I fixed this with:
import json
json.loads(my_data)
You are sending one parameter incorrectly; it should be a dictionary object
:
Wrong: func(a=r)
Correct: func(a={'x':y})
In my case, my get_context_data
in one of my views was returning return render(self.request, 'es_connection_error.html', {'error':error});
in a try/catch block instead of returning context
Here is how I encountered this error in Django and fixed it:
urlpatterns = [path('home/', views.home, 'home'),]
urlpatterns = [path('home/', views.home, name='home'),]