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pythonjsonflaskpython-zipfile

How to send a dynamically generate zipfile to the client


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

Solution

  • 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.