how to unit test file upload in django

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how to unit test file upload in django

2023-02-15 19:29| 来源: 网络整理| 查看: 265

Question :

In my django app, I have a view which accomplishes file upload.The core snippet is like this

... if (requesthod == 'POST'): if request.FILES.has_key('file'): file = request.FILES['file'] with open(settings.destfolder+'/%s' % file.name, 'wb+') as dest: for chunk in file.chunks(): dest.write(chunk)

I would like to unit test the view.I am planning to test the happy path as well as the fail path..ie,the case where the request.FILES has no key ‘file’ , case where request.FILES['file'] has None..

How do I set up the post data for the happy path?Can somebody tell me?

Answer :

I used to do the same with open('some_file.txt') as fp: but then I needed images, videos and other real files in the repo and also I was testing a part of a Django core component that is well tested, so currently this is what I have been doing:

from django.core.files.uploadedfile import SimpleUploadedFile def test_upload_video(self): video = SimpleUploadedFile("file.mp4", "file_content", content_type="video/mp4") self.client.post(reverse('app:some_view'), {'video': video}) # some important assertions ...

In Python 3.5+ you need to use bytes object instead of str. Change "file_content" to b"file_content"

It’s been working fine, SimpleUploadedFile creates an InMemoryFile that behaves like a regular upload and you can pick the name, content and content type.

Answer 2:

From Django docs on Client.post:

Submitting files is a special case. To POST a file, you need only

provide the file field name as a key, and a file handle to the file

you wish to upload as a value. For example:

c = Client() with open('wishlist.doc') as fp: c.post('/customers/wishes/', {'name': 'fred', 'attachment': fp})

Answer 3:

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.

“unit” testing means to test just one “unit” of functionality. So,

if you want to test that view you’d be testing the view, and the file

system, ergo, not really unit test. To make this point more clear. If

you run that test, and the view works fine, but you don’t have

permissions to save that file, your test would fail because of that.

Other important thing is test speed. If you’re doing something like

TDD the speed of execution of your tests is really important.

Accessing any I/O is not a good idea.

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.

Answer 4:

I do something like this for my own event related application but you should have more than enough code to get on with your own use case

import tempfile, csv, os class UploadPaperTest(TestCase): def generate_file(self): try: myfile = open('test.csv', 'wb') wr = csv.writer(myfile) wr.writerow(('Paper ID','Paper Title', 'Authors')) wr.writerow(('1','Title1', 'Author1')) wr.writerow(('2','Title2', 'Author2')) wr.writerow(('3','Title3', 'Author3')) finally: myfile.close() return myfile def setUp(self): self.user = create_fuser() self.profile = ProfileFactory(user=self.user) self.event = EventFactory() self.client = Client() self.module = ModuleFactory() self.event_module = EventModule.objects.get_or_create(event=self.event, module=self.module)[0] add_to_admin(self.event, self.user) def test_paper_upload(self): response = self.client.login(username=self.user.email, password='foz') self.assertTrue(response) myfile = self.generate_file() file_path = myfile.name f = open(file_path, "r") url = reverse('registration_upload_papers', args=[self.event.slug]) # post wrong data type post_data = {'uploaded_file': i} response = self.client.post(url, post_data) self.assertContains(response, 'File type is not supported.') post_data['uploaded_file'] = f response = self.client.post(url, post_data) import_file = SubmissionImportFile.objects.all()[0] self.assertEqual(SubmissionImportFile.objects.all().count(), 1) #self.assertEqual(import_file.uploaded_file.name, 'files/registration/{0}'.format(file_path)) os.remove(myfile.name) file_path = import_file.uploaded_file.path os.remove(file_path)

Answer 5:

I did something like that :

from django.core.files.uploadedfile import SimpleUploadedFile from django.test import TestCase from django.core.urlresolvers import reverse from django.core.files import File from django.utils.six import BytesIO from .forms import UploadImageForm from PIL import Image from io import StringIO def create_image(storage, filename, size=(100, 100), image_mode='RGB', image_format='PNG'): """ Generate a test image, returning the filename that it was saved as. If ``storage`` is ``None``, the BytesIO containing the image data will be passed instead. """ data = BytesIO() Image.new(image_mode, size).save(data, image_format) data.seek(0) if not storage: return data image_file = ContentFile(data.read()) return storage.save(filename, image_file) class UploadImageTests(TestCase): def setUp(self): super(UploadImageTests, self).setUp() def test_valid_form(self): ''' valid post data should redirect The expected behavior is to show the image ''' url = reverse('image') avatar = create_image(None, 'avatar.png') avatar_file = SimpleUploadedFile('front.png', avatar.getvalue()) data = {'image': avatar_file} response = self.client.post(url, data, follow=True) image_src = response.context.get('image_src') self.assertEquals(response.status_code, 200) self.assertTrue(image_src) self.assertTemplateUsed('content_upload/result_image.html')

create_image function will create image so you don’t need to give static path of image.

Note : You can update code as per you code.

This code for Python 3.6.

Answer 6:from rest_framework.test import force_authenticate from rest_framework.test import APIRequestFactory factory = APIRequestFactory() user = User.objects.get(username='#####') view = .as_view() with open('.pdf', 'rb') as fp: request=factory.post('',{'file_name':fp}) force_authenticate(request, user) response = view(request)

Answer 7:

As mentioned in Django’s official documentation:

Submitting files is a special case. To POST a file, you need only provide the file field name as a key, and a file handle to the file you wish to upload as a value. For example:

c = Client() with open('wishlist.doc') as fp: c.post('/customers/wishes/', {'name': 'fred', 'attachment': fp}) More Information: How to check if the file is passed as an argument to some function?

While testing, sometimes we want to make sure that the file is passed as an argument to some function.

e.g.

... class AnyView(CreateView): ... def post(self, request, *args, **kwargs): attachment = request.FILES['attachment'] # pass the file as an argument my_function(attachment) ...

In tests, use Python’s mock something like this:

# Mock 'my_function' and then check the following: response = do_a_post_request() self.assertEqual(mock_my_function.call_count, 1) self.assertEqual( mock_my_function.call_args, call(response.wsgi_request.FILES['attachment']), )

Answer 8:

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.

Answer 9:

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

Answer 10:from django.test import Client from requests import Response client = Client() with open(template_path, 'rb') as f: file = SimpleUploadedFile('Name of the django file', f.read()) response: Response = client.post(url, format='multipart', data={'file': file})

Hope this helps.

Answer 11:

Very handy solution with mock

from django.test import TestCase, override_settings #use your own client request factory from my_framework.test import APIClient from django.core.files import File import tempfile from pathlib import Path import mock image_mock = mock.MagicMock(spec=File) image_mock.name = 'image.png' # or smt else class MyTest(TestCase): # I assume we want to put this file in storage # so to avoid putting garbage in our MEDIA_ROOT # we're using temporary storage for test purposes @override_settings(MEDIA_ROOT=Path(tempfile.gettempdir())) def test_send_file(self): client = APIClient() client.post( '/endpoint/' {'file':image_mock}, format="multipart" )

Answer 12:

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'} )

Answer 13:

I am using django rest framework and I had to test the upload of multiple files.

I finally get it by using format="multipart" in my APIClient.post request.

from rest_framework.test import APIClient ... self.client = APIClient() with open('./photo.jpg', 'rb') as fp: resp = self.client.post('/upload/', {'images': [fp]}, format="multipart")

Answer 14:

I am using GraphQL, upload for test:

with open('test.jpg', 'rb') as fp: response = self.client.execute(query, variables, data={'image': [fp]})

code in class mutation

@classmethod def mutate(cls, root, info, **kwargs): if image := info.context.FILES.get("image", None): kwargs["image"] = image TestingMainModel.objects.get_or_create( id=kwargs["id"], defaults=kwargs )

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