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
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Answer
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.