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.
我正在寻找一种方法将zipfile发送到客户端,该方法是从请求响应生成的。在此示例中,我将一个JSON字符串发送到url,该URL返回已转换的JSON字符串的zip文件。
@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
但是我的zip文件是空的,而烧瓶返回的错误是
Debugging middleware caught exception in streamed response at a point where response
headers were already sent
1 个解决方案
#1
6
You should return the file directly, not a ZipFile()
object:
您应该直接返回文件,而不是ZipFile()对象:
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.
你收到的回复确实是一个zip文件,但没有必要让Python解析它并给你解压缩内容,而Flask肯定不知道如何处理该对象。
#1
6
You should return the file directly, not a ZipFile()
object:
您应该直接返回文件,而不是ZipFile()对象:
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.
你收到的回复确实是一个zip文件,但没有必要让Python解析它并给你解压缩内容,而Flask肯定不知道如何处理该对象。