ValueError: too many values to unpack (expected 2) Django - django

Question

Not sure why but my code is getting the following error:
ValueError: too many values to unpack (expected 2)
Here is my models.py:
class UserList(models.Model):
list_name = models.CharField(max_length=255)
user = models.ForeignKey(User, on_delete=models.CASCADE)
def __str__(self):
return self.list_name
Here is my views.py
def otherUserList(request):
userName = request.GET.get('userName', None)
print(userName)
qs = UserList.objects.filter(user__username=userName)
return qs
Here is the traceback:
Internal Server Error: /electra/otheruserlist/
Traceback (most recent call last):
File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/core/handlers/exception.py", line 47, in inner
response = get_response(request)
File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/utils/deprecation.py", line 116, in __call__
response = self.process_response(request, response)
File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/middleware/clickjacking.py", line 26, in process_response
if response.get('X-Frame-Options') is not None:
File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/query.py", line 418, in get
clone = self._chain() if self.query.combinator else self.filter(*args, **kwargs)
File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/query.py", line 942, in filter
return self._filter_or_exclude(False, *args, **kwargs)
File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/query.py", line 962, in _filter_or_exclude
clone._filter_or_exclude_inplace(negate, *args, **kwargs)
File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/query.py", line 969, in _filter_or_exclude_inplace
self._query.add_q(Q(*args, **kwargs))
File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/sql/query.py", line 1358, in add_q
clause, _ = self._add_q(q_object, self.used_aliases)
File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/sql/query.py", line 1377, in _add_q
child_clause, needed_inner = self.build_filter(
File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/sql/query.py", line 1255, in build_filter
arg, value = filter_expr
ValueError: too many values to unpack (expected 2)
[02/Sep/2021 05:38:43] "GET /electra/otheruserlist/?userName=alice HTTP/1.1" 500 94227
I'd be very grateful for any assistance.

Answer 1

Your view is returning a QuerySet, but that does not make much sense: a view should return a HttpResponse. You can for example render a template, convert it to JSON, etc.
We can for example render a template with:
from django.shortcuts import render
def otherUserList(request):
userName = request.GET.get('userName', None)
qs = UserList.objects.filter(user__username=userName)
return render(request, 'some-template.html', {'lists': qs})
Or we can for example return a JsonResponse wit the list_name`s of user the user with the given username with:
from django.http import JsonField
def otherUserList(request):
userName = request.GET.get('userName', None)
qs = UserList.objects.filter(user__username=userName)
return JsonResponse({'listnames': [list.name for list in qs]})

Answer 2

The exception comes from a Django internal middleware, since it's trying to process your returned qs as a response.
You'll need to return a response, not just a queryset, e.g. this simple example to return a list of user IDs.
from django.http import JsonResponse
def otherUserList(request):
userName = request.GET.get('userName', None)
print(userName)
qs = UserList.objects.filter(user__username=userName)
return JsonResponse({"ids": [user.id for user in qs]})