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<p>我是python的新用户。我不知道为什么,但请求总是引发InvalidURL异常:</p>
<pre><code>>>> import requests
>>> r = requests.get('https://www.google.es/')
</code></pre>
<p>输出:</p>
<pre><code>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/
</code></pre>
<p>此错误与我给出的url无关。我该怎么处理</p>
<p>Python的请求版本为3.7.7和2.23.0</p>
<p>致以最良好的祝愿</p>