python - Python方式不打印错误消息

Question

我试图在调用库时抑制日志中的错误/警告。假设我有此代码

        try:
            kazoo_client.start()
        except:
            pass

这正在调用Zookeeper客户端，该客户端抛出一些异常，该异常会冒泡，现在我不希望在我调用 kazoo_client.start()时在日志中发出警告/错误，有没有办法在您调用客户端时抑制这种情况

Answer 1

假设python 2.7.17

试试这个方法:

import sys, StringIO

def funky() :
 "1" + 1 # This should raise an error

sys.stderr = StringIO.StringIO()
funky() # this should call the funky function

您的代码应如下所示:

import sys, StringIO

# import kazoo somehere around here

sys.stderr = StringIO.StringIO()
kazoo_client.start()

最后是 Python 3 示例:

import sys
from io import StringIO
# import kazoo somehere around here

sys.stderr = StringIO()
kazoo_client.start()