please excuse me for my ugly english ;-)
Imagine this very simple model :
class Photo(models.Model):
image = models.ImageField('Label', upload_to='path/')
I would like to create a Photo from an image URL (i.e., not by hand in the django admin site).
I think that I need to do something like this :
from myapp.models import Photo
import urllib
img_url = 'http://www.site.com/image.jpg'
img = urllib.urlopen(img_url)
# Here I need to retrieve the image (as the same way that if I put it in an input from admin site)
photo = Photo.objects.create(image=image)
I hope that I've well explained the problem, if not tell me.
Thank you :)
Edit :
This may work but I don't know how to convert content to a django File :
from urlparse import urlparse
import urllib2
from django.core.files import File
photo = Photo()
img_url = 'http://i.ytimg.com/vi/GPpN5YUNDeI/default.jpg'
name = urlparse(img_url).path.split('/')[-1]
content = urllib2.urlopen(img_url).read()
# problem: content must be an instance of File
photo.image.save(name, content, save=True)
I just created http://www.djangosnippets.org/snippets/1890/ for this same problem. The code is similar to pithyless' answer above except it uses urllib2.urlopen because urllib.urlretrieve doesn't perform any error handling by default so it's easy to get the contents of a 404/500 page instead of what you needed. You can create callback function & custom URLOpener subclass but I found it easier just to create my own temp file like this:
from django.core.files import File
from django.core.files.temp import NamedTemporaryFile
img_temp = NamedTemporaryFile(delete=True)
img_temp.write(urllib2.urlopen(url).read())
img_temp.flush()
im.file.save(img_filename, File(img_temp))
from myapp.models import Photo
import urllib
from urlparse import urlparse
from django.core.files import File
img_url = 'http://www.site.com/image.jpg'
photo = Photo() # set any other fields, but don't commit to DB (ie. don't save())
name = urlparse(img_url).path.split('/')[-1]
content = urllib.urlretrieve(img_url)
# See also: http://docs.djangoproject.com/en/dev/ref/files/file/
photo.image.save(name, File(open(content[0])), save=True)
Combining what Chris Adams and Stan said and updating things to work on Python 3, if you install Requests you can do something like this:
from urllib.parse import urlparse
import requests
from django.core.files.base import ContentFile
from myapp.models import Photo
img_url = 'http://www.example.com/image.jpg'
name = urlparse(img_url).path.split('/')[-1]
photo = Photo() # set any other fields, but don't commit to DB (ie. don't save())
response = requests.get(img_url)
if response.status_code == 200:
photo.image.save(name, ContentFile(response.content), save=True)
More relevant docs in Django's ContentFile documentation and Requests' file download example.
ImageField is just a string, a path relative to your MEDIA_ROOT setting. Just save the file (you might want to use PIL to check it is an image) and populate the field with its filename.
So it differs from your code in that you need to save the output of your urllib.urlopen to file (inside your media location), work out the path, save that to your model.
I do it this way on Python 3, which should work with simple adaptations on Python 2. This is based on my knowledge that the files I’m retrieving are small. If yours aren’t, I’d probably recommend writing the response out to a file instead of buffering in memory.
BytesIO is needed because Django calls seek() on the file object, and urlopen responses don’t support seeking. You could pass the bytes object returned by read() to Django's ContentFile instead.
from io import BytesIO
from urllib.request import urlopen
from django.core.files import File
# url, filename, model_instance assumed to be provided
response = urlopen(url)
io = BytesIO(response.read())
model_instance.image_field.save(filename, File(io))
Recently I use the following approach within python 3 and Django 3, maybe this might be interesting for others aswell. It is similar to Chris Adams solution but for me it did not work anymore.
import urllib.request
from django.core.files.uploadedfile import SimpleUploadedFile
from urllib.parse import urlparse
from demoapp import models
img_url = 'https://upload.wikimedia.org/wikipedia/commons/f/f7/Stack_Overflow_logo.png'
basename = urlparse(img_url).path.split('/')[-1]
tmpfile, _ = urllib.request.urlretrieve(img_url)
new_image = models.ModelWithImageOrFileField()
new_image.title = 'Foo bar'
new_image.file = SimpleUploadedFile(basename, open(tmpfile, "rb").read())
new_image.save()
Just discovered that you don't have to generate a temporary file:
Stream url content directly from django to minio
I have to store my files in minio and have django docker containers without much disk space and need to download big video files, so this was really helpful to me.
Its been almost 11 years since the question and the most reputed answer has been posted. Thanks To #chris-adams for the response. I am Just reposting the same answer along with the updated packages and support.
#! /usr/bin/python3
# lib/utils.py
import urllib3 # http Request Package.
from typing import Optional
from django.core.files import File # Handle Files in Django
from django.core.files.temp import NamedTemporaryFile # handling temporary files.
def fetch_image(url: str, instance: models.Model, field: str, name: Optional[str]=None):
"""
fetch_image Fetches an image URL and adds it to the model field.
the parameter instance does not need to be a saved instance.
:url: str = A valid image URL.
:instance: django.db.models.Model = Expecting a model with image field or file field.
:field: str = image / file field name as string;
[name:str] = Preferred file name, such as product slug or something.
:return: updated instance as django.db.models.Model, status of updation as bool.
"""
conn = urllib3.PoolManager()
response = conn.request('GET', url)
if response.status <> 200:
print("[X] 404! IMAGE NOT FOUND")
print(f"TraceBack: {url}")
return instance, False
file_obj = NamedTemporaryFile(delete=True)
file_obj.write( response.data )
file_obj.flush()
img_format = url.split('.')[-1]
if name is None:
name = url.split('/')[-1]
if not name.endswith(img_format):
name += f'.{img_format}'
django_file_obj = File(file_obj)
(getattr(instance, field)).save(name, django_file_obj)
return instance, True
Tested with Django==2.2.12 in Python 3.7.5
if __name__ == '__main__':
instance = ProductImage()
url = "https://www.publicdomainpictures.net/pictures/320000/velka/background-image.png"
instance, saved = fetch_image(url, instance, field='banner_image', name='intented-image-slug')
status = ["FAILED! ", "SUCCESS! "][saved]
print(status, instance.banner_image and instance.banner_image.path)
instance.delete()
this is the right and working way
class Product(models.Model):
upload_path = 'media/product'
image = models.ImageField(upload_to=upload_path, null=True, blank=True)
image_url = models.URLField(null=True, blank=True)
def save(self, *args, **kwargs):
if self.image_url:
import urllib, os
from urlparse import urlparse
filename = urlparse(self.image_url).path.split('/')[-1]
urllib.urlretrieve(self.image_url, os.path.join(file_save_dir, filename))
self.image = os.path.join(upload_path, filename)
self.image_url = ''
super(Product, self).save()
This might perhaps be a simple question, but I somehow just can not find the solution. Django offers a lot about uploading file, but how do I do to download a file.
Let's assume we have a button on HTML on uploads/something.txt as a file.
I tried with django.views.static.serve, however what this did it would open a file on webpage.
My question is simple: What is the best and most pythonic way for user of our website to download a file?
You need to read that file.
Serve it using HttpResponse along with proper content type.
Here's some sample code:
content = open("uploads/something.txt").read()
return HttpResponse(content, content_type='text/plain')
This should serve a text file.
But as you described, on some browser, it will not ask to download the file, rather, it would show it in the browser. If you want to show a download prompt, use this:
response = HttpResponse(open("uploads/something.txt", 'rb').read())
response['Content-Type'] = 'text/plain'
response['Content-Disposition'] = 'attachment; filename=DownloadedText.txt'
return response
However, please note that it might be a better idea to serve static contents or uploaded files via nginx or the reverse proxy of your choice. Sending large files through Django might not be the most optimum way of doing that.
import os
from django.conf import settings
from django.http import HttpResponse, Http404
def download(request, path):
file_path = os.path.join(settings.MEDIA_ROOT, path)
if os.path.exists(file_path):
with open(file_path, 'rb') as fh:
response = HttpResponse(fh.read(), content_type="application/vnd.ms-excel")
response['Content-Disposition'] = 'inline; filename=' + os.path.basename(file_path)
return response
raise Http404
Maybe a little late but here is my solution:
def render(self, value):
return format_html('<a href="/media/{0}" download>{0}</a>', value)
I want users to be able to download file from my django web app. I wrote the below codes, but when I download a file, the file will turn into an error file. It won't open and the total size of the file would be 0 bytes.
Models:
class Emov(models.Model):
User=models.ForeignKey(User)
mov_file=models.FileField(upload_to='miiv')
mov_name=models.CharField(max_length=50)
email=models.EmailField() #email of the uploader
download_count=models.PositiveIntegerField(default=0)
#other fields follows
#property
def pretty_name(self):
return "{0}.{1}".format(slugify(self.title),
os.path.splitext(self.mov_name.name)[1])
Views:
def document_view(request,emov_id):
fileload=Emov.objects.get(id=emov_id)
response=HttpResponse()
response["Content-Disposition"]= "attachment; filename={0}".format(fileload.pretty_name)
response['X-Accel-Redirect']="/protected/{0}".format(fileload.mov_name.name)
return response
Nginx
location /protected/ {
internal;
root /C:/Python27/Scripts/env/Scripts/digi/media/miiv/;
}
What am I missing?
Try to do it like this:
fileload = Emov.objects.get(id=emov_id)
filename = fileload.mov_file.name.split('/')[-1]
response = HttpResponse(fileload.mov_file, content_type='text/plain') # your content type
response['Content-Disposition'] = 'attachment; filename=%s' % filename
return response
See:
https://docs.djangoproject.com/en/dev/ref/request-response/#telling-the-browser-to-treat-the-response-as-a-file-attachment
Hope it helps.
EDIT:
FileField.url attribute you should be able to pass a path to your files (like /protected/). https://docs.djangoproject.com/en/dev/ref/models/fields/#filefield-and-fieldfile
I am displaying the list of objects in the html table.
i have the download link in front of every row which i want them to download the linked file.
I have made this function
def make_downloadable_link(path):
#Prepare the form for downloading
wrapper = FileWrapper(open(mypath))
response = HttpResponse(wrapper,'application/pdf')
response['Content-Length'] = os.path.getsize(mypath)
fname = mypath.split('/')[-1]
response['Content-Disposition'] = 'attachment; filename= fname'
return response
This is working fine if i use it for hard coded path in view for single file. But i want to make a generic view so that it works on all the files in the table
I hav the path of the file avaiable in object.path variable but i am confused how can i pass the path object to the downlaod file view. because i want to hide that actual path from the user.
I don't know what to write in the URLs.py file fo that download file view
What would you like to do is get actual file path from object. And as you have said the file path is stored in object.path that makes it easy.
For example:
urls.py
url(r'^download/(?P<object_id>\d+)/$', "yourapp.views.make_downloadable_link", name="downloadable")
In views.py:
def make_downloadable_link(object_id):
# get object from object_id
object = ObjectModel.objects.get(id=object_id)
mypath = object.path
#prepare to serve the file
wrapper = FileWrapper(open(mypath))
response = HttpResponse(wrapper,'application/pdf')
response['Content-Length'] = os.path.getsize(mypath)
fname = mypath.split('/')[-1]
response['Content-Disposition'] = 'attachment; filename= fname'
return response
I have a such problem - I am using Python 2.6 / Django 1.3 and I need to accept as POST variable with key 'f', which contains a binary data. After that, I need to save data in a file.
POST
T$topX$objectsX$versionY$archiverО©ҐR$0О©ҐО©ҐО©Ґull_=<---------------------- content of file -------------------->О©ҐО©Ґ_NSKeyedArchive(258:=CО©ҐО©Ґ
Code
from django.core.files.storage import default_storage
from django.core.files.base import ContentFile
def save(request):
upload_file = request.POST['f']
save_path = default_storage.save('%s%s' % (save_dir, filename),
ContentFile(upload_file))
When I am trying to do
nano /tmp/myfile.zip
It returns data like
T^#^#^#$^#^#^#t^#^#^#o^#^#^#p^#^#^#X^#^#^#$^#^#^#o^#^#^#b^#^#^#j^#^#^#e^#^#^#c^#^#^#t^#^#^#s^#^#^#X^#^#^#$^#^#^#v^#^#^#e^#^#^#r^#^#^#s^#^#^#i^#^#$
When its done, I am going to read saved file
def read(request):
user_file = default_storage.open(file_path).read()
file_name = get_filename(file_path)
response = HttpResponse(user_file, content_type = 'text/plain',
mimetype = 'application/force-download')
response['Content-Disposition'] = 'attachment; filename=%s' % file_name
response['Content-Length'] = default_storage.size(file_path)
return response
In case, when I am writing
print user_file
It returns a correct data, but when I am returning a HttpResponse it has a different data from a source
It would probably be easier, and more memory efficient if you just save the data into a file, and like #keckse said, let a browser stream it. Django is very inefficient in streaming data. It will all depend on the size of the data. If you want to stream it with django anyways, it can be done like this:
from django.http import HttpResponse
import os.path
import mimetypes
def stream(request, document, type=None):
doc = Document.objects.get(pk=document)
fsock = open(doc.file.path,"r")
file_name = os.path.basename(doc.file.path)
mime_type_guess = mimetypes.guess_type(file_name)
if mime_type_guess is not None:
response = HttpResponse(fsock, mimetype=mime_type_guess[0])
response['Content-Disposition'] = 'attachment; filename=' + file_name
return response
In your case you might want to set the mime type manually, you can try out application/octet-stream too. The mainpassing iterators difference is that you pass the "string" from file.read(), instead of the handle to the file directly. Please note: if you use read(), you will be loading the whole file into memory.
More on passing iterators to HttpResonse. And I might be wrong, but I think you can drop the content-type.