CODEWITHSUNDEEP
pythonzipstringio
nabucosound
nabucosound

Reputation: 1293

Function to create in-memory zip file and return as http response

I am avoiding the creation of files on disk, this is what I have got so far:

def get_zip(request):
    import zipfile, StringIO
    i = open('picture.jpg', 'rb').read()
    o = StringIO.StringIO()
    zf = zipfile.ZipFile(o, mode='w')
    zf.writestr('picture.jpg', i)
    zf.close()
    o.seek(0)
    response = HttpResponse(o.read())
    o.close()
    response['Content-Type'] = 'application/octet-stream'
    response['Content-Disposition'] = "attachment; filename=\"picture.zip\""
    return response

Do you think is correct-elegant-pythonic enough? Any better way to do it?

Thanks!

Upvotes: 6

Views: 7150

Answers (2)

Lukáš Lalinský
Lukáš Lalinský

Reputation: 41306

For StringIO you should generally use o.getvalue() to get the result. Also, if you want to add a normal file to the zip file, you can use zf.write('picture.jpg'). You don't need to manually read it.

Upvotes: 12

S.Lott
S.Lott

Reputation: 391952

Avoiding disk files can slow your server to a crawl, but it will certainly work.

You'll exhaust memory if you serve too many of these requests concurrently.

Upvotes: 4

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