Flask: how to send a dynamically generate zipfile to the client

问题

I am looking for a way to send a zipfile to the client that is generated from a requests response. In this example, I send a JSON string to a url which returns a zip file of the converted JSON string.

@app.route('/sendZip', methods=['POST'])
def sendZip():
    content = '{"type": "Point", "coordinates": [-105.01621, 39.57422]}'
    data = {'json' : content}
    r = requests.post('http://ogre.adc4gis.com/convertJson', data = data)
    if r.status_code == 200:
        zipDoc = zipfile.ZipFile(io.BytesIO(r.content))
        return Response(zipDoc,
                mimetype='application/zip',
                headers={'Content-Disposition':'attachment;filename=zones.zip'})

But my zip file is empty and the error returned by flask is

Debugging middleware caught exception in streamed response at a point where response 
headers were already sent

回答1:

You should return the file directly, not a ZipFile() object:

r = requests.post('http://ogre.adc4gis.com/convertJson', data = data)
if r.status_code == 200:
    return Response(r.content,
            mimetype='application/zip',
            headers={'Content-Disposition':'attachment;filename=zones.zip'})

The response you receive is indeed a zipfile, but there is no point in having Python parse it and give you unzipped contents, and Flask certainly doesn't know what to do with that object.

来源：https://stackoverflow.com/questions/26513542/flask-how-to-send-a-dynamically-generate-zipfile-to-the-client