How to download a file in Django | Making a large file downloadable from a URL in Django | How to serve Large files in Django | Downloading a file from the Django project directory

Sometimes we have to serve a file which may be a large file from our project directory. Suppose you want to transfer a file from your project directory to your friend or to some other person/server/platform. One way is to download the file and then upload it to another platform or give it to the person manually, which is not a good approach.

This is a simple thing in Django. You can serve any file from any directory (i.e. inside the project directory) using a URL in Django. Just follow the steps shown below and you will find that some line of code will help you to do that.

If the file is binary:

Add the following URL in URL patterns.

#In urls.py
path('download/',views.download,name="download"),

Now set the function in views.py i.e. download() as mapped in urls.py

#In views.py
from django.http import FileResponse
import os

def download(request):
  file_path = os.path.join(BASE_DIR, 'myfile.zip')
  response = FileResponse(open(file_path, 'rb'))
  return response

So now if someone visits "www.example.com/download" the downloading of file i.e "myfile.zip" will start.
Note that if the file is not binary then you have to use 'r' instead of 'rb' in your code.

If the file is a text/UTF-8 file

If you only want to show the text content of the file then you can use the following code:

#In views.py
from django.http import FileResponse
import os

def download(request):
  file_path = os.path.join(BASE_DIR, 'requirement.txt')
  f = open(file_path, 'r')
  file_content = f.read()
  f.close()
  return HttpResponse(file_content, content_type="text/javascript")

In this case, the text content of the file will be server in the browser or client device.
Now, if the file is stored in database models/fields then you can do something like this:

#In views.py
from django.http import FileResponse
import os

def download(request):
  obj=mymodel.objects.get(id=someid)
  response = FileResponse(open(obj.fileField.path, 'rb'))
  return response