Django: simulate HTTP requests in shell

How I simulate requests from the python command line is:

• Use the excellent requests library
• Use the django reverse function

A simple way of simulating requests is:

>>> from django.urls import reverse
>>> import requests
>>> r = requests.get(reverse('app.views.your_view'))
>>> r.text
(prints output)
>>> r.status_code
200

Update: be sure to launch the django shell (via manage.py shell), not a classic python shell.

Update 2: For Django <1.10, change the first line to

from django.core.urlresolvers import reverse 

You can use RequestFactory, which allows

• inserting a user into the request

• inserting an uploaded file into the request

• sending specific parameters to the view

and does not require the additional dependency of using requests.

Note that you have to specify both the URL and the view class, so it takes an extra line of code than using requests.

from django.test import RequestFactory

request_factory = RequestFactory()
my_url = '/my_full/url/here'  # Replace with your URL -- or use reverse
my_request = request_factory.get(my_url)
response = MyClasBasedView.as_view()(my_request)  # Replace with your view
response.render()
print(response)

To set the user of the request, do something like my_request.user = User.objects.get(id=123) before getting the response.

To send parameters to a class-based view, do something like response = MyClasBasedView.as_view()(my_request, parameter_1, parameter_2)

Extended Example

Here's an example of using RequestFactory with these things in combination

• HTTP POST (to url url, functional view view, and a data dictionary post_data)

• uploading a single file (path file_path, name file_name, and form field value file_key)

• assigning a user to the request (user)

• passing on kwargs dictionary from the url (url_kwargs)

SimpleUploadedFile helps format the file in a way that is valid for forms.

from django.core.files.uploadedfile import SimpleUploadedFile
from django.test import RequestFactory

request = RequestFactory().post(url, post_data)
with open(file_path, 'rb') as file_ptr:
    request.FILES[file_key] = SimpleUploadedFile(file_name, file_ptr.read())
    file_ptr.seek(0)  # resets the file pointer after the read
    if user:
        request.user = user
    response = view(request, **url_kwargs)

Using RequestFactory from a Python shell

RequestFactory names your server "testserver" by default, which can cause a problem if you're not using it inside test code. You'll see an error like:

DisallowedHost: Invalid HTTP_HOST header: 'testserver'. You may need to add 'testserver' to ALLOWED_HOSTS.

This workaround from @boatcoder's comment shows how to override the default server name to "localhost":

request_factory = RequestFactory(**{"SERVER_NAME": "localhost", "wsgi.url_scheme":"https"}).