pyspider中Exception: HTTP 599: <url> malformed问题

Question

[E 170810 14:07:02 base_handler:203] HTTP 599: <url> malformed

Traceback (most recent call last):
  File "/usr/local/lib/python2.7/dist-packages/pyspider/libs/base_handler.py", line 196, in run_task
    result = self._run_task(task, response)
  File "/usr/local/lib/python2.7/dist-packages/pyspider/libs/base_handler.py", line 175, in _run_task
    response.raise_for_status()
  File "/usr/local/lib/python2.7/dist-packages/pyspider/libs/response.py", line 172, in raise_for_status
    six.reraise(Exception, Exception(self.error), Traceback.from_string(self.traceback).as_traceback())
  File "/usr/local/lib/python2.7/dist-packages/pyspider/fetcher/tornado_fetcher.py", line 378, in http_fetch
    response = yield gen.maybe_future(self.http_client.fetch(request))
  File "/usr/local/lib/python2.7/dist-packages/tornado/httpclient.py", line 102, in fetch
    self._async_client.fetch, request, **kwargs))
  File "/usr/local/lib/python2.7/dist-packages/tornado/ioloop.py", line 458, in run_sync
    return future_cell[0].result()
  File "/usr/local/lib/python2.7/dist-packages/tornado/concurrent.py", line 238, in result
    raise_exc_info(self._exc_info)
  File "<string>", line 3, in raise_exc_info
Exception: HTTP 599: <url> malformed

Answer 1

异常说你的URL有问题，请给出你的URL