In my django app, I have a view which accomplishes file upload.The core snippet is like this
...
if (request.method == \'POST\'):
if request.FILES.has_
from rest_framework.test import force_authenticate
from rest_framework.test import APIRequestFactory
factory = APIRequestFactory()
user = User.objects.get(username='#####')
view = <your_view_name>.as_view()
with open('<file_name>.pdf', 'rb') as fp:
request=factory.post('<url_path>',{'file_name':fp})
force_authenticate(request, user)
response = view(request)
if you want to add other data with file upload then follow the below method
file = open('path/to/file.txt', 'r', encoding='utf-8')
data = {
'file_name_to_receive_on_backend': file,
'param1': 1,
'param2': 2,
.
.
}
response = self.client.post("/url/to/view", data, format='multipart')`
The only file_name_to_receive_on_backend
will be received as a file other params received normally as post paramas.
I recommend you to take a look at Django RequestFactory. It's the best way to mock data provided in the request.
Said that, I found several flaws in your code.
So, I recommend you to refactor your view to use a function like:
def upload_file_to_location(request, location=None): # Can use the default configured
And do some mocking on that. You can use Python Mock.
PS: You could also use Django Test Client But that would mean that you're adding another thing more to test, because that client make use of Sessions, middlewares, etc. Nothing similar to Unit Testing.
In Django 1.7 there's an issue with the TestCase wich can be resolved by using open(filepath, 'rb') but when using the test client we have no control over it. I think it's probably best to ensure file.read() returns always bytes.
source: https://code.djangoproject.com/ticket/23912, by KevinEtienne
Without rb option, a TypeError is raised:
TypeError: sequence item 4: expected bytes, bytearray, or an object with the buffer interface, str found
I am using Python==3.8.2 , Django==3.0.4, djangorestframework==3.11.0
I tried self.client.post
but got a Resolver404
exception.
Following worked for me:
import requests
upload_url='www.some.com/oaisjdoasjd' # your url to upload
with open('/home/xyz/video1.webm', 'rb') as video_file:
# if it was a text file we would perhaps do
# file = video_file.read()
response_upload = requests.put(
upload_url,
data=video_file,
headers={'content-type': 'video/webm'}
)