我正在尝试使用 python 请求获得响应。但是我面临 SSL 握手错误。我尝试了此处发布的许多解决方案,但似乎没有任何效果。请帮忙。提前致谢
这是代码:
import requests
url = "https://androidappsapk.co/download/com.facebook.katana"
requests.get(url, verify = False)
这是我遇到的错误:
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/local/lib/python2.7/site-packages/requests/api.py", line 55, in get
return request('get', url, **kwargs)
File "/usr/local/lib/python2.7/site-packages/requests/api.py", line 44, in request
return session.request(method=method, url=url, **kwargs)
File "/usr/local/lib/python2.7/site-packages/requests/sessions.py", line 456, in request
resp = self.send(prep, **send_kwargs)
File "/usr/local/lib/python2.7/site-packages/requests/sessions.py", line 559, in send
r = adapter.send(request, **kwargs)
File "/usr/local/lib/python2.7/site-packages/requests/adapters.py", line 382, in send
raise SSLError(e, request=request)
requests.exceptions.SSLError: [SSL: SSLV3_ALERT_HANDSHAKE_FAILURE] sslv3 alert handshake failure (_ssl.c:590)
附言- 我正在使用 Python 2.7.11
最佳答案
我将 OpenSSL 更新到最新版本,现在工作正常:)
关于python - Python 中的 SSL 握手错误 (_ssl.c :590),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44644849/