在Django中使用Python ZipStream获取损坏的zip

Question

I am a little new to Django so please bear with me.
I'm using zipstream from here and have a Django view that returns a zip file of all file attachments which are all hosted on Amazon S3. But the zip files are all coming up as corrupt when I download them, that is, I can't open them. I have tried verifying the files with unzip -t but the errors are not very helpful.

file_paths = [fa.file.url for fa in file_attachments.all()]

zf = zipstream.ZipFile(mode='w', compression=zipstream.ZIP_DEFLATED)

zip_subdir = "Attachments-%s" % (request_id)

for file_path in file_paths:
    file_dir, file_name = os.path.split(file_path)

    zf.writestr(file_name, urllib.urlopen(file_path).read())

zip_filename = "%s.zip" % (zip_subdir)

response = StreamingHttpResponse(zf, mimetype='application/zip')
response['Content-Disposition'] = \
    'attachment; filename={}'.format(zip_filename)
return response

Any ideas?

Solved it.

s = StringIO.StringIO()
with zipstream.ZipFile(s, mode='w', compression=zipstream.ZIP_DEFLATED) as zf:
    #could fail on url open.
    for file_path in file_paths:
        file_dir, file_name = os.path.split(file_path)

        try:
            file_contents = urllib.urlopen(file_path).read()
            zf.writestr(file_name, file_contents)
        except IOError: #connection cannot be made
            logging.error()

response = StreamingHttpResponse(s.getvalue(), mimetype='application/octet-stream')
response['Content-Disposition'] = \
    'attachment; filename={}'.format("%s" % (request_id))
return response

Answer 1

You should close the ZipFile when you're done writing to it. Otherwise, to quote the documentation , "essential records will not be written" until you do.

The cleanest way to do it is using the with statement:

with zipstream.ZipFile(mode='w', compression=zipstream.ZIP_DEFLATED) as zf:
    # ...write to zf...