Python请求总是抛出异常,与所写的url无关

问题描述 投票:0回答:1

抱歉,因为我是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/

我写什么网址都没关系。您能帮我吗?

Python的版本是3.7.7和2.23.0用于请求。

最诚挚的问候。

python python-requests python-3.7
1个回答
0
投票
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)。

错误不是由requests引起的,而是在安装urllib3时安装的requests 2.21.0+(新版本)中的问题。为了避免这种情况,请尝试更新urllib3

python -m pip install --upgrade urllib3

或安装requests v2.21.0

pip uninstall requests # to remove current version pip install requests==2.21.0

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