Вот мой код:
class PublishPhotosHandler(webapp.RequestHandler):
for argument in files_arguments:
taskqueue.add(url='/upload', params={'key': key})
class UploadWorker(webapp.RequestHandler):
def post(self):
key = self.request.get('key')
result = urlfetch.fetch(...)
# how to return there an error, so the task will be retried?