将blob上载到appstore：查询参数以获取upload u

Question

我正在使用App Engine for Python : Uploading a BLOB中的BLOB上载示例 唯一的区别是，由于我已经在使用Django REST框架，所以我决定使用它而不是webapp2来实现示例代码中的MainHandler类（我没有使用表单）

class PdfUploadUrl(APIView):

    def get(self, request, *args, **kwargs):
        ##
        # Fetch the URL to which we upload the PDF
        ##
        log.debug('PdfUploadUrl: request.GET = %s:%s', type(request.GET), request.GET)
        upload_url = blobstore.create_upload_url('/pdfservice/upload')
        log.debug('PdfUploadUrl: upload_url = %s : %s', type(upload_url), upload_url)
        return Response(dict(upload_url=str(upload_url)))

当我这样称呼它的时候

^{pr2}$

答案是

^{3}$

现在，当我把请求改为

curl -i -G http://<my-app>.appspot.com/pdf/uploadurl?xyz=bc

那么，答案是

{"upload_url": "http://<my-app>.appspot.com/_ah/upload/?xyz=bc/AMmfu6b4YEpDMuiFI494dY39yCCAehiAykQ2hy8fWEo1GS7kuQ6BKG9zR8TJhfDzlWbH6Fvc9TZWNBGTca7PY3DxY8f8-KKiKzIbS2k09CeMXBGVhfJE_EgGH0l_P3pL1ndVmL-6lGOF/ALBNUaYAAAAAUwbjrZJHTdW6rGDJZxLCuwbq46bw3Bk7/"}

这实际上是一个不可用的URL。为什么查询参数会扰乱对的调用

     blobstore.create_upload_url()