CODEWITHSUNDEEP
pythondjango
Mirage
Mirage

Reputation: 31548

How can I pass kwargs in URL in django

In the django doc the url function is like this

url(regex, view, kwargs=None, name=None, prefix='')

I have this

url(r'^download/template/(?P<object_id>\d+)/$', views.myview().myfunction,model=models.userModel, name="sample")

This is my view

class myview(TemplateView):

    def myfunction(self,request, object_id, **kwargs):
        model = kwargs['model']

I get this error

url() got an unexpected keyword argument 'model'

Upvotes: 17

Views: 38508

Answers (4)

deepak Kumar
deepak Kumar

Reputation: 11

This is how our url should look like:

url(regex, view, kwargs=None, name=None, prefix='')

The main part here is that this kwargs argument should be a dictionary. This is the problem with your code.

So, instead of passing model=models.userModel, you should pass it in the form of a dictionary in your url like this:

kwargs=dict(model=models.userModel)

Because if you look carefully in your views.py code, you will notice that you were trying to put kwargs['model'] (the value stored in the kwargs dictionary with the model key) into the model variable.

class myview(TemplateView):

    def myfunction(self,request, object_id, **kwargs):
        model = kwargs['model']

But since you were not passing any dictionary, you were not able to access the kwargs dictionary.

Upvotes: 1

jpic
jpic

Reputation: 33420

This:

url(r'^download/template/(?P<object_id>\d+)/$', views.myview().myfunction,model=models.userModel, name="sample")

Should be:

url(r'^download/template/(?P<object_id>\d+)/$', views.myview.as_view(model=models.userModel), name="sample")

See docs

Your current implementation is not thread safe. For example:

from django import http
from django.contrib.auth.models import User
from django.views import generic


class YourView(generic.TemplateView):
    def __init__(self):
        self.foo = None

    def your_func(self, request, object_id, **kwargs):
        print 'Foo', self.foo
        self.foo = 'bar'
        return http.HttpResponse('foo')



urlpatterns = patterns('test_app.views',
    url(r'^download/template/(?P<object_id>\d+)/$', YourView().your_func,
        kwargs=dict(model=User), name="sample"),
)

Would you expect it to print 'Foo None' ? Well be careful cause the instance is shared across requests:

Django version 1.4.2, using settings 'django_test.settings'
Development server is running at http://127.0.0.1:8000/
Quit the server with CONTROL-C.
Foo None
[03/Dec/2012 08:14:31] "GET /test_app/download/template/3/ HTTP/1.1" 200 3
Foo bar
[03/Dec/2012 08:14:32] "GET /test_app/download/template/3/ HTTP/1.1" 200 3

So, when it's not thread safe, you can't assume that it will be in a clean state when the request starts - unlike when using as_view().

Upvotes: 8

Jacob Valenta
Jacob Valenta

Reputation: 6769

I believe you would have the same functionality (and avoid threading issues) if you did this in your views.py

from django.views.generic import TemplateView
from .models import userModel

class myview(TemplateView):
    def myfunction(self, request, object_id, *args, **kwargs):
        model = userModel
        # ... Do something with it

Upvotes: 0

Martijn Pieters
Martijn Pieters

Reputation: 1121864

You are trying to pass in a model keyword argument to the url() function; you need to pass in a kwargs argument instead (it takes a dictionary):

url(r'^download/template/(?P<object_id>\d+)/$', views.myview().myfunction, 
    kwargs=dict(model=models.userModel), name="sample")

Upvotes: 20

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