Django 'function' object has no attribute 'objects'

Question

My apps allow you to like a picture then it redirects you to the same page

I get then error when I try to like a picture I can create a like objects with the shell prompt but why I get this error? thank for helping me

AttributeError at /like/3/
function' object has no attribute 'objects'Request Method: GET 
Request URL: http://127.0.0.1:8000/like/3/ 

Exception Value: 'function' object has no attribute 'objects' 
Traceback:
File "C:\Python26\Lib\site-packages\django\core\handlers\base.py" in get_response
111.                         response = callback(request, *callback_args, **callback_kwargs)
File "C:\o\mysite\pet\views.py" in Like
195.     new_like, created = Like.objects.get_or_create(user=request.user, picture_id=picture_id)

This is parts of my views.py

def Like(request,picture_id):
    pid = picture_id
    new_like, created = Like.objects.get_or_create(user=request.user, picture_id=picture_id)
    p = Picture.objects.get(pk=pid)
    if created:
        HttpResponseRedirect(reverse('world:url_name'))
    else:
        HttpResponseRedirect(reverse('world:url_name'))

My URLconf:

     url(

Parts of my model: r'^like/(?P\d+)/$', 'pet.views.Like', name = 'Like' ), My boat.html

 {% if picture %}
 <ul>
    {% for pet in picture %}
    <li><b>description</b> = {{ pet.description}}<br/>
        {% if pet.image %}
 <li>
    <a href ="{% url world:Like pet.id %}">
        <img src= "{{ pet.image.url }}" style="cursor:pointer">
    </a>
 <li>
        {% endif %}
 {% endfor %}
 </ul>
 {% endif %}
 <a href="{% url world:PictureCreator %}">Add Pictures to your board</a><br/>

My models.py

class Picture(models.Model):
    user = models.ForeignKey(User)
    board = models.ForeignKey(Board,blank=False,null=False)
    image = models.FileField(upload_to="images/",blank=True)
    description = models.TextField()
    is_primary = models.BooleanField(default=False)
    def __unicode__(self):
        return self.description

class Like(models.Model):
    user = models.ForeignKey(User)
    picture = models.ForeignKey(Picture)
    created = models.DateTimeField(auto_now_add=True)

This message appeared to me because I put a @api_view() decorator on top of a user model (don't me ask how!). After removing it, everything worked properly again. — claudius
– claudius, Commented Jun 5, 2023 at 23:09

dm03514 · Accepted Answer · 2013-03-16 14:04:00Z

28

your view function name is defined as Like and your model is named Like

you define Like as a function so when you go to access Like.objects python does not see your model Like but the function Like

you could rename your view function

url(r'^like/(?P\d+)/$', 'pet.views.change_name_no_conflict', name = 'Like' )


def change_name_no_conflict(request,picture_id):
  pass

answered Mar 16, 2013 at 14:04

dm03514

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Comments

Sandeep G · Accepted Answer · 2016-06-04 03:20:27Z

7

Model name and view name shouldn't be same.

answered Jun 4, 2016 at 3:20

Sandeep G

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Darwin · Accepted Answer · 2021-05-18 19:43:12Z

1

This is because your function name in view.py and model name in models.py are the same. You can change the function name or model name.

Another solution is:

from .models import modelname as modelname2

def modelname(request):
    obj_list_ads = modelname2.objects.all()

in this code function name is modelname.

answered May 18, 2021 at 19:43

Darwin

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DevB2F · Accepted Answer · 2017-07-04 17:37:13Z

0

I got this same error when doing

 from django.contrib.auth import get_user_model
 User = get_user_model

adding () solved the error:

 from django.contrib.auth import get_user_model
 User = get_user_model()

answered Jul 4, 2017 at 17:37

DevB2F

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Ankit Anand · Accepted Answer · 2019-09-18 03:38:28Z

0

function name and model name does depend on name, function name should be same as url name we define url in urls.py model name depend on function data member, its means as for example when we take data from user and save in database then we call that object from its model name ex= u_data = registration() ,this is used for user data seve and define that fun where we send data to save means target url and in its related view.

answered Sep 18, 2019 at 3:38

Ankit Anand

1

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Code on the Rocks · Accepted Answer · 2022-06-21 01:37:21Z

0

I hit this error because my view's query set looked like this:

class IngredientList(generics.ListAPIView):
  queryset = Ingredient.objects.all # This line is wrong
  serializer_class = IngredientSerializer

I fixed it by adding a set of parenthesis to the queryset line:

class IngredientList(generics.ListAPIView):
      queryset = Ingredient.objects.all() # Fixed
      serializer_class = IngredientSerializer

answered Jun 21, 2022 at 1:37

Code on the Rocks

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4b0 · Accepted Answer · 2020-04-05 09:36:36Z

-1

Define object:

objects = models.Manager()

Your error will solve!

edited Apr 5, 2020 at 9:36

4b0

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answered Apr 2, 2020 at 22:27

Aman Bothra

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