如何处理requests.exceptions.InvalidURL：无法在python中解析？

Question

我是python的新用户。我不知道为什么，但是请求总是抛出InvalidURL异常：

>>> import requests
>>> r = requests.get('https://www.google.es/')

输出：

Traceback (most recent call last):
  File "/usr/local/lib/python3.7/dist-packages/requests/models.py", line 380, in prepare_url
    scheme, auth, host, port, path, query, fragment = parse_url(url)
  File "/usr/lib/python3/dist-packages/urllib3/util/url.py", line 392, in parse_url
    return six.raise_from(LocationParseError(source_url), None)
  File "<string>", line 3, in raise_from
urllib3.exceptions.LocationParseError: Failed to parse: https://www.google.es/

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/local/lib/python3.7/dist-packages/requests/api.py", line 76, in get
    return request('get', url, params=params, **kwargs)
  File "/usr/local/lib/python3.7/dist-packages/requests/api.py", line 61, in request
    return session.request(method=method, url=url, **kwargs)
  File "/usr/local/lib/python3.7/dist-packages/requests/sessions.py", line 516, in request
    prep = self.prepare_request(req)
  File "/usr/local/lib/python3.7/dist-packages/requests/sessions.py", line 459, in prepare_request
    hooks=merge_hooks(request.hooks, self.hooks),
  File "/usr/local/lib/python3.7/dist-packages/requests/models.py", line 314, in prepare
    self.prepare_url(url, params)
  File "/usr/local/lib/python3.7/dist-packages/requests/models.py", line 382, in prepare_url
    raise InvalidURL(*e.args)
requests.exceptions.InvalidURL: Failed to parse: https://www.google.es/

此错误与我提供的网址无关。我该如何处理？

Python的版本是3.7.7和2.23.0用于请求。

最诚挚的问候。

Answer 1

由于urllib3的新版本，您遇到了错误（某些用户往往会遇到此问题）。