Ubuntu服务器上的Scrapy Splash：得到了一个意想不到的关键字参数'encoding'

Question

我正在使用的scrapy plash在本地计算机上工作正常，但是当我在Ubuntu Server上使用它时，它会返回此错误。这是为什么？是由低记忆引起的吗？

  File "/usr/local/lib64/python2.7/site-packages/twisted/internet/defer.py", line 1299, in _inlineCallbacks
    result = g.send(result)
  File "/usr/local/lib/python2.7/site-packages/scrapy/core/downloader/middleware.py", line 53, in process_response
    spider=spider)
  File "/usr/local/lib/python2.7/site-packages/scrapy_splash/middleware.py", line 387, in process_response
    response = self._change_response_class(request, response)
  File "/usr/local/lib/python2.7/site-packages/scrapy_splash/middleware.py", line 402, in _change_response_class
    response = response.replace(cls=respcls, request=request)
  File "/usr/local/lib/python2.7/site-packages/scrapy/http/response/text.py", line 50, in replace
    return Response.replace(self, *args, **kwargs)
  File "/usr/local/lib/python2.7/site-packages/scrapy/http/response/__init__.py", line 79, in replace
    return cls(*args, **kwargs)
  File "/usr/local/lib/python2.7/site-packages/scrapy_splash/response.py", line 33, in __init__
    super(_SplashResponseMixin, self).__init__(url, *args, **kwargs)
TypeError: __init__() got an unexpected keyword argument 'encoding'

更新

它仅在使用localhost作为 SPLASH_URL

Answer 1

TypeError： init （）有一个意外的关键字参数'编码'

看来，您将某个地方启动（实例创建可能）传递给一个额外的参数"编码"。尝试调查。

Answer 2

我通过使用确切的URL来解决它：

SPLASH_URL = 'http://therealip:8050'

任何Localhost解决方案均行不通。我认为这是废品飞溅的错误。

update

事实证明，如果我关闭crawlera，错误也消失了。但是随后它重现了另一个错误。最好不要在配置中使用本地主机。