I try to download an image from my django website. I do it like this:
def file_download(request, filename):
from django.core.servers.basehttp import FileWrapper
import mimetypes
import settings
import os
filepath = os.path.join(settings.MEDIA_ROOT, filename)
wrapper = FileWrapper(open(filepath))
content_type = mimetypes.guess_type(filepath)[0]
response = HttpResponse(wrapper, mimetype='content_type')
response['Content-Disposition'] = "attachment; filename=%s" % filename
return response
However, it doesn't work for images (I tries jpg files), but do work for txt files. Why?
Probably you need to open the file in binary mode:
wrapper = FileWrapper(open(filepath, 'rb'))
Related
I am trying to download a file that is stored in the media folder and it seems as it only works when I am writing the full path as:
file_location = '/home/user/user.pythonanywhere.com/media/'
In the settings page on pythonanywhere, the 'media url' points to the same directory. So I don't really understand why I just can't write /media/ instead for the full path.
Has it something to do with my settings.py file? I have these lines there:
MEDIA_ROOT= os.path.join(BASE_DIR, 'media')
MEDIA_URL="/media/"
Do I need to have these lines at all?
This is my download function:
class RequestFile(APIView):
def get(self, request):
#Returns the last uploaded file
#Remember if deleting files, only database record is deleted. The file must be deleted
obj = FileUploads.objects.all().order_by('-id')
file_location = '/home/user/user.pythonanywhere.com/media/' + str(obj[0].lastpkg)
x = file_location.find("/")
filename = file_location[x+1:]
print(file_location)
print(filename)
try:
with open(file_location, 'rb') as f:
filex_data = f.read()
#print("filex_data= ", filex_data)
# sending response
response = HttpResponse(filex_data, content_type='application/octet-stream')
response['Content-Disposition'] = 'attachment; filename=' + filename
except IOError:
# handle file not exist case here
response = HttpResponseNotFound('File not exist')
return response
If I go to the url that points to this class I will get the download.
From Django documentation
>>> car = Car.objects.get(name="57 Chevy")
>>> car.photo
<ImageFieldFile: cars/chevy.jpg>
>>> car.photo.name
'cars/chevy.jpg'
>>> car.photo.path
'/media/cars/chevy.jpg'
>>> car.photo.url
'http://media.example.com/cars/chevy.jpg'
So in your case something like
obj = FileUploads.objects.first()
try:
with open(obj.name_of_file_attribute.path, 'rb') as f:
I have a Django app that will be placed in a Docker container.
The app prepares data in Dataframe format. I would like to allow the user to download the data to his/her local drive as excel file.
I have used df.to_excel in the past, but this won't work in this case.
Please advise best way to do this.
As of pandas-0.17, you can let Django write to a BytesIO directly, like:
from django.http import HttpResponse
from io import BytesIO
def some_view(request):
with BytesIO() as b:
# Use the StringIO object as the filehandle.
writer = pd.ExcelWriter(b, engine='xlsxwriter')
df.to_excel(writer, sheet_name='Sheet1')
writer.save()
# Set up the Http response.
filename = 'django_simple.xlsx'
response = HttpResponse(
b.getvalue(),
content_type='application/vnd.openxmlformats-officedocument.spreadsheetml.sheet'
)
response['Content-Disposition'] = 'attachment; filename=%s' % filename
return response
You might need to install an Excel writer module (like xlsxwriter, or openpyxl).
I think it can be even simpler and more concise these days. You can just pass the http response directly to the Excel writer. The following works for me:
from django.http import HttpResponse
import pandas as pd
# df = CREATE YOUR OWN DATAFRAME
response = HttpResponse(content_type='application/xlsx')
response['Content-Disposition'] = f'attachment; filename="FILENAME.xlsx"'
with pd.ExcelWriter(response) as writer:
df.to_excel(writer, sheet_name='SHEET NAME')
return response
I have developed an API (Python 3.5, Django 1.10, DRF 3.4.2) that uploads a video file to my media path when I request it from my UI. That part is working fine. I try to write a test for this feature but cannot get it to run successfully.
#views.py
import os
from rest_framework import views, parsers, response
from django.conf import settings
class FileUploadView(views.APIView):
parser_classes = (parsers.FileUploadParser,)
def put(self, request, filename):
file = request.data['file']
handle_uploaded_file(file, filename)
return response.Response(status=204)
def handle_uploaded_file(file, filename):
dir_name = settings.MEDIA_ROOT + '/scene/' + filename + '/cam1'
new_filename = 'orig.mp4'
if not os.path.exists(dir_name):
os.makedirs(dir_name)
file_path = os.path.join(dir_name, new_filename)
with open(file_path, 'wb+') as destination:
for chunk in file.chunks():
destination.write(chunk)
and
#test.py
import tempfile
import os
from django.test import TestCase
from django.conf import settings
from django.core.files import File
from django.core.files.uploadedfile import SimpleUploadedFile
from rest_framework.test import APIRequestFactory
from myapp.views import FileUploadView
class UploadVideoTestCase(TestCase):
def setUp(self):
settings.MEDIA_ROOT = tempfile.mkdtemp(suffix=None, prefix=None, dir=None)
def test_video_uploaded(self):
"""Video uploaded"""
filename = 'vid'
file = File(open('media/testfiles/vid.mp4', 'rb'))
uploaded_file = SimpleUploadedFile(filename, file.read(), 'video')
factory = APIRequestFactory()
request = factory.put('file_upload/'+filename,
{'file': uploaded_file}, format='multipart')
view = FileUploadView.as_view()
response = view(request, filename)
print(response)
dir_name = settings.MEDIA_ROOT + '/scene/' + filename + '/cam1'
new_filename = 'orig.mp4'
file_path = os.path.join(dir_name, new_filename)
self.assertTrue(os.path.exists(file_path))
In this test, I need to use an existing video file ('media/testfiles/vid.mp4') and upload it since I need to test some processings on the video data after: that's why I reset the MEDIA_ROOT using mkdtemp.
The test fails since the file is not uploaded. In the def put of my views.py, when I print request I get <rest_framework.request.Request object at 0x10f25f048> and when I print request.data I get nothing. But if I remove the FileUploadParser in my view and use request = factory.put('file_upload/' + filename, {'filename': filename}, format="multipart") in my test, I get <QueryDict: {'filename': ['vid']}> when I print request.data.
So my conclusion is that the request I generate with APIRequestFactory is incorrect. The FileUploadParseris not able to retrieve the raw file from it.
Hence my question: How to generate a file upload (test) request with Django REST Framework's APIRequestFactory?
Several people have asked questions close to this one on SO but I had no success with the proposed answers.
Any help on that matter will be much appreciated!
It's alright now! Switching from APIRequestFactory to APIClient, I managed to have my test running.
My new test.py:
import os
import tempfile
from django.conf import settings
from django.core.files import File
from django.core.files.uploadedfile import SimpleUploadedFile
from django.urls import reverse
from rest_framework.test import APITestCase, APIClient
from django.contrib.auth.models import User
class UploadVideoTestCase(APITestCase):
def setUp(self):
settings.MEDIA_ROOT = tempfile.mkdtemp()
User.objects.create_user('michel')
def test_video_uploaded(self):
"""Video uploaded"""
filename = 'vid'
file = File(open('media/testfiles/vid.mp4', 'rb'))
uploaded_file = SimpleUploadedFile(filename, file.read(),
content_type='multipart/form-data')
client = APIClient()
user = User.objects.get(username='michel')
client.force_authenticate(user=user)
url = reverse('file_upload:upload_view', kwargs={'filename': filename})
client.put(url, {'file': uploaded_file}, format='multipart')
dir_name = settings.MEDIA_ROOT + '/scene/' + filename + '/cam1'
new_filename = 'orig.mp4'
file_path = os.path.join(dir_name, new_filename)
self.assertTrue(os.path.exists(file_path))
Below, testing file upload using APIRequestFactory as requested (and ModelViewSet).
from rest_framework.test import APIRequestFactory, APITestCase
from my_project.api.views import MyViewSet
from io import BytesIO
class MyTestCase(APITestCase):
def setUp(self):
fd = BytesIO(b'Test File content') # in-memory file to upload
fd.seek(0) # not needed here, but to remember after writing to fd
reqfactory = APIRequestFactory() # initialize in setUp if used by more tests
view = MyViewSet({'post': 'create'}) # for ViewSet {action:method} needed, for View, not.
request = factory.post('/api/new_file/',
{
"title": 'test file',
"fits_file": self.fd,
},
format='multipart') # multipart is default, but for clarification that not json
response = view(request)
response.render()
self.assertEqual(response.status_code, 201)
Note that there is no authorization for clarity, as with: 'DEFAULT_PERMISSION_CLASSES': ['rest_framework.permissions.AllowAny'].
If I run:
127.0.0.1:8000/document/1/preview
this pdf file is downloaded.
I need display it in HTML(preview with print function). How to do it?
views.py
from xhtml2pdf import pisa
from django.template.loader import render_to_string
from datetime import datetime
import StringIO
def pdf_report(request, did):
d_instance = get_object_or_404(MyObject, pk=did, user=request.user)
contents = render_to_string('pdf_preview.html', {'object':d_instance})
response = HttpResponse(mimetype='application/pdf')
response['Content-Disposition'] = 'attachment; filename=answer_%s.pdf' % (f_date,)
result = StringIO.StringIO()
pdf = pisa.pisaDocument(StringIO.StringIO(contents.encode('utf-8')), result, show_error_as_pdf=True, encoding='UTF-8')
if not pdf.err:
response.write(result.getvalue())
result.close()
return response
urls.py
(r'^document/(?P<did>\d+)/preview/$', 'app.views.pdf_report'),
To make the file open in the browser, use inline content-disposition.
response['Content-Disposition'] = 'inline; filename=answer_%s.pdf' % f_date
Attached is the code which downloads a file from browser using django 1.3 and Apache 2.2 with mod_xsendfile
#login_required
def sendfile(request, productid):
path = settings.RESOURCES_DIR
filepath = os.path.join('C:/workspace/y/src/y/media/audio/','sleep_away.mp3')
print "filepath",filepath
filename = 'sleep_away.mp3' # Select your file here.
print "Within sendfile size", os.path.getsize(filepath)
wrapper = FileWrapper(open(filepath,'r'))
content_type = mimetypes.guess_type(filename)[0]
response = HttpResponse(wrapper, content_type = content_type)
print "Within wrapper"
from django.utils.encoding import smart_str
response['X-Sendfile'] = smart_str(filepath)
response['Content-Length'] = os.path.getsize(filepath)
from django.utils.encoding import smart_str
response['Content-Disposition'] = 'attachment; filename=%s/' % smart_str(filename)
return response
The console shows the following filesize which is the right size
Within sendfile size 4842585
But when I download/save the file it shows 107 KB...i.e 109,787 bytes.Where am I going wrong. Why isnt it downloading the complete file?
I consider your new to django or python. Try to put the import statements at the beginning of the method. Once imported it can be used through the method no need import every time you use. In windows you should use "rb" (read binary) to serve anything other than text files. Try not to use variable names that might conflict with method names or other keywords of the language. Your method should be like this
#login_required
def sendfile(request, productid):
from django.utils.encoding import smart_str
##set path and filename
resource_path = settings.RESOURCES_DIR # resource dir ie /workspace/y/src/y/media
filename = "sleep_away.mp3" #file to be served
##add it to os.path
filepath = os.path.join(resource_path,"audio",filename)
print "complete file path: ", filepath
##filewrapper to server in size of 8kb each until whole file is served
file_wrapper = FileWrapper(file(filepath,'rb')) ##windows needs rb (read binary) for non text files
##get file mimetype
file_mimetype = mimetypes.guess_type(filepath)
##create response with file_mimetype and file_wrapper
response = HttpResponse(content_type=file_mimetype, file_wrapper)
##set X-sendfile header with filepath
response['X-Sendfile'] = filepath ##no need for smart_str here.
##get filesize
print "sendfile size", os.stat(filepath).st_size
response['Content-Length'] = os.stat(filepath).st_size ##set content length
response['Content-Disposition'] = 'attachment; filename=%s/' % smart_str(filename) ##set disposition
return response ## all done, hurray!! return response :)
Hope that helps
You could have a look at the django-private-files project. Haven't tested it myself, but it looks promissing.
link to the docs --> http://readthedocs.org/docs/django-private-files/en/latest/usage.html
cheers