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To send a file response to a user in a Django application following processing, follow these steps:

1. Process the file (e.g. convert format or perform calculations)
2. Store the processed file in a directory or save it as a temporary file
3. Create a HttpResponse object with the processed file as its content
4. Add headers to the response object with the content type and disposition
5. Return the response object

Here's an example code snippet that demonstrates this procedure:

import os
from django.http import HttpResponse

def process_file(request):
    # process the file here and save it to a directory or a temporary file
    processed_file_path = '/path/to/processed/file'

    # create a response object with the processed file as its content
    with open(processed_file_path, 'rb') as file:
        response = HttpResponse(file.read())

    # add headers for content type and disposition
    content_type = 'application/octet-stream'
    response['Content-Type'] = content_type
    response['Content-Disposition'] = f'attachment; filename={os.path.basename(processed_file_path)}'
    response['Content-Length'] = os.path.getsize(processed_file_path)

    return response

In this example, process_file takes a request object as input and performs some processing on a file, which is then saved at processed_file_path. A HttpResponse object is then created with the processed file as its content, and headers are added to specify the content type, disposition and length. The final step is to return the response object.