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I'm currently using openPYXL in order to open a template file within Django

    module_dir = os.path.dirname(__file__)  # get current directory
file_path = os.path.join(module_dir, fileName)
username = request.user.username
workbook = load_workbook(file_path)
worksheet = workbook.active

The file is then edited, and saved under a different name(This works fine, I can open the created file and it contains the information desired), however what I'm struggling with is serving this file to the user, I've tried various techniques such as shown below

workbook.save('EvalofSelf1.xlsx')
response = HttpResponse()
file_path = os.path.join(os.path.dirname(os.path.realpath(__name__)), 'EvalofSelf1.xlsx')
response['X-Sendfile'] = file_path
response['Content-Type'] = 'mimetype/submimetype'
response['Content-Disposition'] = 'attachment; filename=%s.xlsx' % 'DownloadedEval'

All of which serve a file as requested, but the file contains no actual data, is 0kb in size and unopenable, how can I serve up the created file from my Django project directory, retaining all information stored within it?

kxdan
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1 Answers1

5

You're not ever putting the file contents into the response, so naturally it is 0 bytes. X-Sendfile is for a completely different purpose - when you're redirecting to a static server - and needs a URL, not a file path, anyway.

file_path = os.path.join(os.path.dirname(os.path.realpath(__name__)), 'EvalofSelf1.xlsx')
response = HttpResponse(open(file_path, 'rb').read())
response['Content-Type'] = 'mimetype/submimetype'
response['Content-Disposition'] = 'attachment; filename=DownloadedEval.xlsx'
Daniel Roseman
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