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如何处理 requests.exceptions.InvalidURL:在 python 中解析失败?

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

>>> import requests
>>> r = requests.get('https://www.google.es/')
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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, …
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python python-requests python-3.7

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