单击上下文菜单项后验证当前URL

Question

我正在编写一个python脚本，右键单击一个文件，单击一个上下文菜单项。点击打开一个网页。我必须验证网页的网址。如何将控制权从操作系统转移到浏览器并获取当前URL。在

Answer 1

你为什么不打电话请求.get（url）并检查响应代码。另一个选择是打电话请求.head（url）

>>> import requests
>>> url1 = 'http://example.com'
>>> url2 = 'http://sdsdsdsdsdss.com'
>>> r = requests.head(url1)
>>> r.status_code
200
>>> r = requests.head(url2, timeout=5)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.7/dist-packages/requests/api.py", line 77, in head
    return request('head', url, **kwargs)
  File "/usr/lib/python2.7/dist-packages/requests/api.py", line 44, in request
    return session.request(method=method, url=url, **kwargs)
  File "/usr/lib/python2.7/dist-packages/requests/sessions.py", line 383, in request
    resp = self.send(prep, **send_kwargs)
  File "/usr/lib/python2.7/dist-packages/requests/sessions.py", line 486, in send
    r = adapter.send(request, **kwargs)
  File "/usr/lib/python2.7/dist-packages/requests/adapters.py", line 387, in send
    raise Timeout(e)
requests.exceptions.Timeout: (<urllib3.connectionpool.HTTPConnectionPool object at 0x7f4e77635950>, 'Connection to sdsdsdsdsdss.com timed out. (connect timeout=5)')
>>> 

你需要处理这个异常。有关请求模块的详细信息：http://docs.python-requests.org/en/latest/

如果您真的需要打开web浏览器，可以使用这个库：https://docs.python.org/2/library/webbrowser.html