I work on a small Django app and get an error tells me, super(type, obj): obj must be an instance or subtype of type. I get it from the views.py file after introducing the function get_object_or_404. The views.py file provided below,
from django.shortcuts import render, get_object_or_404
from django.http import HttpResponse, HttpResponseRedirect
from django.views import View
from .models import URL
# function based view
def redirect_view(request, shortcode=None, *args, **kwargs):
obj = get_object_or_404(URL, shortcode=shortcode)
return HttpResponse("Hello World, the shortcode is {shortcode}".format(shortcode = obj.url))
# class based view
class ShortenerView(View):
def get(self, request, shortcode=None, *args, **kwargs):
obj = get_object_or_404(URL, shortcode=shortcode)
return HttpResponse("Hello World 1, the shortcode is {shortcode}".format(shortcode = obj.url))
def post(self, request, *args, **kwargs):
return HttpResponse()
the full error message is here,
TypeError at /b/p6jzbp/
super(type, obj): obj must be an instance or subtype of type
Request Method: GET
Request URL: http://127.0.0.1:8000/b/p6jzbp/
Django Version: 1.11
Exception Type: TypeError
Exception Value:
super(type, obj): obj must be an instance or subtype of type
Exception Location: /Users/Chaklader/Documents/Projects/UrlShortener/src/shortener/models.py in all, line 18
The line 18 in the models.py is qs_main = super(URL, self).all(*args, **kwargs) and the models.py file is here,
# will look for the "SHORTCODE_MAX" in the settings and
# if not found, will put the value of 15 there
SHORTCODE_MAX = getattr(settings, "SHORTCODE_MAX", 15)
class UrlManager(models.Manager):
def all(self, *args, **kwargs):
qs_main = super(URL, self).all(*args, **kwargs)
qs = qs_main.filter(active = True)
return qs
def refresh_shortcodes(self, items = None):
qs = URL.objects.filter(id__gte=1)
new_codes = 0
if items is not None and isinstance(items, int):
qs = qs.order_by('-id')[:items]
for q in qs:
q.shortcode = create_shortcode(q)
print (q.id, " ", q.shortcode)
q.save()
new_codes += 1
return "# new codes created {id}".format(id = new_codes)
class URL(models.Model):
url = models.CharField(max_length = 220, )
shortcode = models.CharField(max_length = SHORTCODE_MAX, blank = True, unique = True)
updated = models.DateTimeField(auto_now = True)
timestamp = models.DateTimeField(auto_now_add = True)
active = models.BooleanField(default = True)
objects = UrlManager()
def save(self, *args, **kwargs):
if self.shortcode is None or self.shortcode == "":
self.shortcode = create_shortcode(self)
super(URL, self).save(*args, **kwargs)
def __str__(self):
return str(self.url)
def __unicode__(self):
return str(self.url)
# class Meta:
# ordering = '-id'
Can someone explain the the reason of error to me and how to solve it? I'm open to provide more informations IF required.
You should call super using the UrlManager class as first argument not the URL model. super cannot called be with an unrelated class/type:
From the docs,
super(type[, object-or-type]): Return a proxy object that delegates method calls to a parent or sibling class of type.
So you cannot do:
>>> class D:
... pass
...
>>> class C:
... def __init__(self):
... super(D, self).__init__()
...
>>> C()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 3, in __init__
TypeError: super(type, obj): obj must be an instance or subtype of type
You should do:
qs_main = super(UrlManager, self).all(*args, **kwargs)
Or in Python 3:
qs_main = super().all(*args, **kwargs)
Another interesting way is if a merge of branches has duplicated the class, so that in the file you have two definitions for the same name, e.g.
class A(Foo):
def __init__(self):
super(A, self).__init__()
#...
class A(Foo):
def __init__(self):
super(A, self).__init__()
#...
If you try to create an instance from a static reference to the first definition of A, once it tries to call super, inside the __init__ method, A will refer to the second definition of A, since it has been overwritten. The solution - ofcourse - is to remove the duplicate definition of the class, so it doesn't get overwritten.
This may seem like something that would never happen, but it just happened to me, when I wasn't paying close enough attention to the merge of two branches. My tests failed with the error message described in the question, so I thought I'd leave my findings here, even though it doesn't exactly answer the specific question.
Another way this error can occur is when you reload the module with the class in a Jupiter notebook.
Easy solution is to restart the kernel.
http://thomas-cokelaer.info/blog/2011/09/382/
Check out @Mike W's answer for more detail.
Elaborating in @Oğuz Şerbetci's answer, in python3 (not necessary only in Jupyter), when there is the need to reload a library, for example we have class Parent and class Child defined as
class Parent(object):
def __init__(self):
# do something
class Child(Parent):
def __init__(self):
super(Child, self).__init__(self)
then if you do this
import library.Child
reload(library)
Child()
you will get TypeError: super(type, obj): obj must be an instance or subtype of type, the solution is just to re import the class after the reload
import library.Child
reload(library)
import library.Child
Child()
For Jupyter only
You can get his issue in because reload logic have some bugs (issue)
Here is a simple solution/workaround that works for me until issue is not fixed
1001xx at the bottom of the file which you call in the cell The best solution that I have found for this problem is only available using python 3. You then don't need to specify the arguments of "super", then you won't have the error any more writing your class like this :
class D:
pass
class C(D):
def __init__(self):
super().__init__()# no arguments given to super()
This error also pops out when you simply do not instantiate child class , and try to call a method on a class itself, like in :
class Parent:
def method():
pass
class Child(Parent):
def method():
super().method()
P = Parent()
C = Child
C.method()