django RequestFactory file upload - django

Question

I try to create a request, using RequestFactory and post with file, but I don't get request.FILES.
from django.test.client import RequestFactory
from django.core.files import temp as tempfile
tdir = tempfile.gettempdir()
file = tempfile.NamedTemporaryFile(suffix=".file", dir=tdir)
file.write(b'a' * (2 ** 24))
file.seek(0)
post_data = {'file': file}
request = self.factory.post('/', post_data)
print request.FILES # get an empty request.FILES : <MultiValueDict: {}>
How can I get request.FILES with my file ?

Answer 1

If you open the file first and then assign request.FILES to the open file object you can access your file.
request = self.factory.post('/')
with open(file, 'r') as f:
request.FILES['file'] = f
request.FILES['file'].read()
Now you can access request.FILES like you normally would. Remember that when you leave the open block request.FILES will be a closed file object.

Answer 2

I made a few tweaks to #Einstein 's answer to get it to work for a test that saves the uploaded file in S3:
request = request_factory.post('/')
with open('my_absolute_file_path', 'rb') as f:
request.FILES['my_file_upload_form_field'] = f
request.FILES['my_file_upload_form_field'].read()
f.seek(0)
...
Without opening the file as 'rb' I was getting some unusual encoding errors with the file data
Without f.seek(0) the file that I uploaded to S3 was zero bytes

Answer 3

You need to provide proper content type, proper file object before updating your FILES.
from django.core.files.uploadedfile import File
# Let django know we are uploading files by stating content type
content_type = "multipart/form-data; boundary=------------------------1493314174182091246926147632"
request = self.factory.post('/', content_type=content_type)
# Create file object that contain both `size` and `name` attributes
my_file = File(open("/path/to/file", "rb"))
# Update FILES dictionary to include our new file
request.FILES.update({"field_name": my_file})
the boundary=------------------------1493314174182091246926147632 is part of the multipart form type. I copied it from a POST request done by my webbrowser.

Answer 4

All the previous answers didn't work for me. This seems to be an alternative solution:
from django.core.files.uploadedfile import SimpleUploadedFile
with open(file, "rb") as f:
file_upload = SimpleUploadedFile("file", f.read(), content_type="text/html")
data = {
"file" : file_upload
}
request = request_factory.post("/api/whatever", data=data, format='multipart')

Answer 5

Be sure that 'file' is really the name of your file input field in your form.
I got that error when it was not (use name, not id_name)