I want to load a default django page. Nothing fancy. However, the error I get, hints at an id that is incorrectly set.
"Field 'id' expected a number but got 'zoekboek'."
The confusing things here (I am a django beginner, so I wouldn't be surprised if this is not confusing at all for you):
the path for this page in the urls.py is not asking for an id.
the view is not querying anything yet (I found some posts that had similar errors,
but related to a filter).
the debug info points to another view that indeed is requesting an id.
when I add a slash at the beginning of the path, the error is gone!
The code
urls.py
urlpatterns = [
path('', views.scholen, name='scholen'),
path('<school_id>', views.school_detail, name='school_detail'),
path('<school_id>/<groep_id>', views.school_groep, name='school_groep'),
path('<school_id>/<groep_id>/<UserProfile_id>', views.leerling_page, name='leerling_page'),
path('zoekboek', views.zoekboek, name='zoekboek'),
]
views.py
from django.shortcuts import render, redirect, reverse, get_object_or_404
from books.models import Book, Rating
from .models import School, Groep
from profiles.models import UserProfile, Hobby, Sport
from django.contrib.auth.models import User
# Create your views here.
def scholen(request):
"""
Homepage for participating
schools.
"""
scholen = School.objects.all()
context = {
'scholen': scholen,
}
return render(request, 'schools/school_landing.html', context)
def school_detail(request, school_id):
"""
Details of individual schools.
"""
school = get_object_or_404(School, pk=school_id)
groep = Groep.objects.filter(school=school)
context = {
'school': school,
'groep': groep,
}
return render(request, 'schools/school_detail.html', context)
def school_groep(request, school_id, groep_id):
"""
Details of groep.
"""
school = get_object_or_404(School, pk=school_id)
groep = get_object_or_404(Groep, pk=groep_id)
a = groep.naam
kinderen = UserProfile.objects.filter(groep=a)
context = {
'school': school,
'groep': groep,
'kinderen': kinderen,
}
return render(request, 'schools/school_groep.html', context)
def leerling_page(request, school_id, groep_id, UserProfile_id):
"""
Personal page of school kids.
"""
profile = get_object_or_404(UserProfile, pk=UserProfile_id)
# If viewer is owner of page, viewer can edit
owner = False
if request.user == profile.user:
owner = True
context = {
'profile': profile,
'owner': owner,
}
return render(request, 'schools/leerling_page.html', context)
def zoekboek(request):
"""
Page for kids to search their favorite book
"""
context = {
}
return render(request, 'schools/zoek_boek.html', context)
Is this enough information?
Simple fix: move path('zoekboek', views.zoekboek, name='zoekboek'), from the last place to the second place in your urls.
Why?
Because Django URLs are resolved using regular expressions; the docs say here in point 3:
Django runs through each URL pattern, in order, and stops at the first one that matches the requested URL, matching against path_info.
Since your URL path path('<school_id>', views.school_detail, name='school_detail'), is very generic, it matches any string including the string zoekboek; so the request to zoekboek falls into the second line in your URL conf and gets routed to the view school_detail() and a school_id is expected for that view.
Suggestion: to make the URL handling easier and so you can order the URL paths however you like, you could change the URL a bit and add a prefix (for example school/) so that not any string matches the URL paths. For example, this schould work:
urlpatterns = [
path('', ...),
path('school/<school_id>', ...),
path('school/<school_id>/<groep_id>', ...),
path('school/<school_id>/<groep_id>/<UserProfile_id>', ...),
path('zoekboek', ...),
]
I created a Flask form where I can upload an image. Then I need to convert that image to base64 string, but I'm always getting the same result.
OUTPUT of my prints:
<FileStorage: '20190925_184412.jpg' ('image/jpeg')>
b''
And the code
from flask import Flask, render_template
from flask_wtf import FlaskForm
from wtforms import FileField
from flask_uploads import configure_uploads, IMAGES, UploadSet
import base64
app = Flask(__name__)
app.config['SECRET_KEY'] = 'thisisasecret'
app.config['UPLOADED_IMAGES_DEST'] = 'uploads/images'
images = UploadSet('images', IMAGES)
configure_uploads(app, images)
class MyForm(FlaskForm):
image = FileField('image')
#app.route('/', methods=['GET', 'POST'])
def index():
form = MyForm()
if form.validate_on_submit():
filename = images.save(form.image.data)
image_string = base64.b64encode(form.image.data.read())
print(form.image.data)
print(image_string)
return f'Filename: { filename }'
return render_template('index.html', form=form)
I think this is due to the way Werkzeug's FileStorage object works. As I menioned in another answer it has a stream attribute; this is of type tempfile.SpooledTemporaryFile so must be re-wound after reading, if you wish to read it again.
In your case this stream attribute is: form.image.data.stream. I suspect this is read once when you call the method images.save.
So the solution should be to rewind that stream, prior to calculating the the b64 string:
if form.validate_on_submit():
filename = images.save(form.image.data) # first read happens here
form.image.data.stream.seek(0)
image_string = base64.b64encode(form.image.data.read())
print(form.image.data)
print(image_string)
return f'Filename: { filename }'
I think the correct way to have instance variables in Flask is by adding users and sessions, but I'm trying to test a concept and I don't want to go through all of that just yet. I'm trying to have a web app load an image into a variable that can then have different image operations performed on it. Obviously you don't want to have to keep performing a list of operations on the image on each new request because that would be horribly inefficient.
Is there a way of having an app.var in Flask that I can access from different routes? I've tried using the global context and Flask's current_app, but I get the impression that's not what they're for.
The code for my blueprint is:
import os
from flask import Flask, url_for, render_template, \
g, send_file, Blueprint
from io import BytesIO
from PIL import Image, ImageDraw, ImageOps
home = Blueprint('home', __name__)
#home.before_request
def before_request():
g.img = None
g.user = None
#home.route('/')
def index():
return render_template('home/index.html')
#home.route('/image')
def image():
if g.img is None:
root = os.path.dirname(os.path.abspath(__file__))
filename = os.path.join(root, '../static/images/lena.jpg')
g.img = Image.open(filename)
img_bytes = BytesIO()
g.img.save(img_bytes, 'jpeg')
img_bytes.seek(0)
return send_file(img_bytes, mimetype='image/jpg')
#home.route('/grayscale', methods=['POST'])
def grayscale():
if g.img:
print('POST grayscale request')
g.img = ImageOps.grayscale(img)
return "Grayscale operation successful"
else:
print('Grayscale called with no image loaded')
return "Grayscale operation failed"
The /image route returns the image correctly, but I'd like to be able to call /grayscale, perform the operation, and be able to make another call to /image and have it return the image from memory without loading it.
You could save a key in your session variable and use that to identify the image in a global dictionary. However this might lead to some trouble if you use multiple Flask application instances. But with one it would be fine. Otherwise you could use Redis when working with multiple workers. I haven't tried the following code but it should show the concept.
from flask import session
import uuid
app.config['SECRET_KEY'] = 'your secret key'
img_dict = {}
#route('/image')
def image():
key = session.get('key')
if key is None:
session['key'] = key = uuid.uuid1()
img_dict[key] = yourimagedata
#home.route('/grayscale', methods=['POST'])
def grayscale():
key = session.get('key')
if key is None:
print('Grayscale called with no image loaded')
return "Grayscale operation failed"
else:
img = img_dict[key]
print('POST grayscale request')
g.img = ImageOps.grayscale(img)
return "Grayscale operation successful"
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_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?
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.
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})
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.
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)
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.
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)
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']),
)
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.
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
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.
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"
)
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'}
)
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")
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
)
The project I'm working on has some data that needs to get passed to every view, so we have a wrapper around render_to_response called master_rtr. Ok.
Now, I need our 404 pages to run through this as well. Per the instructions, I created a custom 404 handler (cleverly called custom_404) that calls master_rtr. Everything looks good, but our tests are failing, because we're receiving back a 200 OK.
So, I'm trying to figure out how to return a 404 status code, instead. There seems to be an HttpResponseNotFound class that's kinda what I want, but I'm not quite sure how to construct all of that nonsense instead of using render_to_response. Or rather, I could probably figure it out, but it seems like their must be an easier way; is there?
The appropriate parts of the code:
def master_rtr(request, template, data = {}):
if request.user.is_authenticated():
# Since we're only grabbing the enrollments to get at the courses,
# doing select_related() will save us from having to hit database for
# every course the user is enrolled in
data['courses'] = \
[e.course for e in \
Enrollment.objects.select_related().filter(user=request.user) \
if e.view]
else:
if "anonCourses" in request.session:
data['courses'] = request.session['anonCourses']
else:
data['courses'] = []
data['THEME'] = settings.THEME
return render_to_response(template, data, context_instance=RequestContext(request))
def custom_404(request):
response = master_rtr(request, '404.html')
response.status_code = 404
return response
The easy way:
def custom_404(request):
response = master_rtr(...)
response.status_code = 404
return response
But I have to ask: why aren't you just using a context processor along with a RequestContext to pass the data to the views?
Just set status_code on the response.
Into your application's views.py add:
# Imports
from django.shortcuts import render
from django.http import HttpResponse
from django.template import Context, loader
##
# Handle 404 Errors
# #param request WSGIRequest list with all HTTP Request
def error404(request):
# 1. Load models for this view
#from idgsupply.models import My404Method
# 2. Generate Content for this view
template = loader.get_template('404.htm')
context = Context({
'message': 'All: %s' % request,
})
# 3. Return Template for this view + Data
return HttpResponse(content=template.render(context), content_type='text/html; charset=utf-8', status=404)
The secret is in the last line: status=404
Hope it helped!