Question

我想使用Python来监控使用HTTPS的网站。问题是网站上的证书无效。我不在乎，我只是想知道该网站正在运行。

我的工作代码如下：

[UIscreen mainscreen].bounds

以URLError结束调用。错误代码为645。

from urllib.request import Request, urlopen
from urllib.error import URLError, HTTPError
req = Request("https://somedomain.com")

try:
    response = urlopen(req)
except HTTPError as e:
    print('server couldn\'t fulfill the request')
    print('error code: ', e.code)
except URLError as e:
        print(e.args)       
else:
    print ('website ok')

所以，我试图将代码645除外。我试过这个：

C:\python>python monitor443.py
(SSLError(1, '[SSL: CERTIFICATE_VERIFY_FAILED] certificate verify failed (_ssl.c:645)'),)

但得到此错误：

from urllib.request import Request, urlopen
from urllib.error import URLError, HTTPError
req = Request("https://somedomain.com")

try:
    response = urlopen(req)
except HTTPError as e:
    print('server couldn\'t fulfill the request')
    print('error code: ', e.code)
except URLError as e:
        if e.code == 645:
            print("ok")
        print(e.args)       
else:
    print ('website ok')

如何添加此例外？

Answer 1

请查看精彩的requests套餐。它可以简化您进行http通信时的生活。请参阅http://requests.readthedocs.io/en/master/。

pip install requests

要跳过证书检查，您可以执行以下操作（请注意verify参数！）：

requests.get('https://kennethreitz.com', verify=False)
<Response [200]>

请参阅完整文档here。

HTH

Answer 2

我无法安装SLL库（egg_info错误）。这就是我最终做的事情

from urllib.request import Request, urlopen
from urllib.error import URLError, HTTPError

def sendEmail(r):    
    #send notification
    print('send notify')

req = Request("https://somedomain.com")

try:
    response = urlopen(req)
except HTTPError as e:
    print('server couldn\'t fulfill the request')
    print('error code: ', e.code)
    sendEmail('server couldn\'t fulfill the request')
except URLError as e:
        theReason=str(e.reason)
        #[SSL: CERTIFICATE_VERIFY_FAILED] certificate verify failed (_ssl.c:645)
        if theReason.find('CERTIFICATE_VERIFY_FAILED') == -1:
            sendEmail(theReason)
        else:
            print('website ok')         
else:
    print('website ok')

python捕获URLError代码

2 个答案: