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
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 am fairly new to Django and my project requires me to prompt user to open a pdf upon clicking a link. I already have the pdf file on my local machine and dont want to recreate it using Reportlab. Is there any way to do it?
I tried
with open("/user/some/directory/somefilename.pdf") as pdf:
response = HttpResponse(pdf, content_type='application/pdf')
response['Content-Disposition'] = 'attachment; filename="somefilename.pdf"'
return response
but it returned 404 page not found as the requested url wasn't in the URLconf of myproject.urls
What am I missing?
In general, when user click "Download", you can:
- If file is not existed:
- Generate pdf file use ReportLab as you did.
- Store generated file to a public dir.
return HttpResponseRedirect(file_url_to_public_dir)
The way that worked for me is by using FileSystemStorage
from django.core.files.storage import FileSystemStorage
from django.http import HttpResponse
fs = FileSystemStorage("/Users/location/where/file/is_saved/")
with fs.open("somefile.pdf") as pdf:
response = HttpResponse(pdf, content_type='application/pdf')
response['Content-Disposition'] = 'attachment; filename="my_pdf.pdf"'
return response
and now its prompting the user to save the file as it normally would!
I have the following script to upload a file unto google drive, using python27. As it is now it will upload a new copy of the file, but I want the existing file updated/overwritten. I can't find help in the Google Drive API references and guides for python. Any suggestions?
from __future__ import print_function
import os
from apiclient.discovery import build
from httplib2 import Http
from oauth2client import file, client, tools
try:
import argparse
flags = argparse.ArgumentParser(parents=[tools.argparser]).parse_args()
except ImportError:
flags = None
# Gain acces to google drive
SCOPES = 'https://www.googleapis.com/auth/drive.file'
store = file.Storage('storage.json')
creds = store.get()
if not creds or creds.invalid:
flow = client.flow_from_clientsecrets('client_secret.json', SCOPES)
creds = tools.run_flow(flow, store, flags) \
if flags else tools.run(flow, store)
DRIVE = build('drive', 'v3', http=creds.authorize(Http()))
#The file that is being uploaded
FILES = (
('all-gm-keys.txt', 'application/vnd.google-apps.document'), #in google doc format
)
#Where the file ends on google drive
for filename, mimeType in FILES:
folder_id = '0B6V-MONTYPYTHONROCKS-lTcXc' #Not the real folder id
metadata = {'name': filename,'parents': [ folder_id ] }
if mimeType:
metadata['mimeType'] = mimeType
res = DRIVE.files().create(body=metadata, media_body=filename).execute()
if res:
print('Uploaded "%s" (%s)' % (filename, res['mimeType']))
I think that you are looking for the update method. Here is a link to the documentation. There is an example on overwriting the file in python.
I think that using the official google client api instead of pure http requests should make your task easier.
from apiclient import errors
from apiclient.http import MediaFileUpload
# ...
def update_file(service, file_id, new_title, new_description, new_mime_type,
new_filename, new_revision):
"""Update an existing file's metadata and content.
Args:
service: Drive API service instance.
file_id: ID of the file to update.
new_title: New title for the file.
new_description: New description for the file.
new_mime_type: New MIME type for the file.
new_filename: Filename of the new content to upload.
new_revision: Whether or not to create a new revision for this file.
Returns:
Updated file metadata if successful, None otherwise.
"""
try:
# First retrieve the file from the API.
file = service.files().get(fileId=file_id).execute()
# File's new metadata.
file['title'] = new_title
file['description'] = new_description
file['mimeType'] = new_mime_type
# File's new content.
media_body = MediaFileUpload(
new_filename, mimetype=new_mime_type, resumable=True)
# Send the request to the API.
updated_file = service.files().update(
fileId=file_id,
body=file,
newRevision=new_revision,
media_body=media_body).execute()
return updated_file
except errors.HttpError, error:
print 'An error occurred: %s' % error
return None
Link the example: https://developers.google.com/drive/api/v2/reference/files/update#examples
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'))
In a view I create new file with:
sys.stdout = open(backup_name, 'w')
call_command('dumpdata')
How can I now return this file to user?
I tried to change mimetype in HttpResponse to 'application/json' but how can I add file content to response?
Or maybe there is other way to return file?
.
I know it's a bit late, but I found this a useful starting point so I thought others could benefit from what I found too.
For a small file, if you place the json file in a template folder, django can find it and you can return it with render_to_response:
return render_to_response(data_file,mimetype='application/json')
I found this to be problematic for large datasets on certain browsers. I would get the error An existing connection was forcibly closed by the remote host. An alternative approach fixed this.
First you must create full path to your file. Use the PROJECT_ROOT variable (defined by PROJECT_ROOT = os.path.abspath(os.path.dirname(__file__)) in settings.py). To access this and the os methods you must import settings, os in views.py. Once you have this file location you can return it using the code below:
backup_path = os.path.join(settings.PROJECT_ROOT, "templates", "json_dumps", "large_file.json")
return HttpResponse(open(backup_path, 'r'),content_type = 'application/json; charset=utf8')
I found this worked well for even very large files.
OK I have it:
response = HttpResponse(open(backup_path, "r"), mimetype='application/json', )
response['Content-Disposition'] = "filename=%s" % backup_name"
After saving file on disc I open it for reading and set file name in response.
Anyone has another idea?
I was trying to return a dictionary as a json file. Here is my solution:
import json
import cStringIO as StringIO
from wsgiref.util import FileWrapper
from django.http import HttpResponse
data_string = json.dumps(data)
json_file = StringIO.StringIO()
json_file.write(data_string)
json_file.seek(0)
wrapper = FileWrapper(json_file)
response = HttpResponse(wrapper, content_type='application/json')
response['Content-Disposition'] = 'attachement; filename=dump.json'
return response
Just copy/link/call the dumpdata code related to model serialization, and dump it directly into the response, so you avoid permission problems and filesystem pollution. Content-disposition and mimetype still applies.
Remember anyway that dumpdata can be a lenghty process, so you are exposed to timeouts.
My final solution is (thanks to saverio):
response = HttpResponse(mimetype='application/json', )
response['Content-Disposition'] = "filename=%s" % backup_name
sys.stdout = response
call_command('dumpdata')