捕获异常后python cmd模块返回提示

问题描述

我有这段代码，我正在使用python 3：

I have this code and I'm using python 3:

import cmd
class myShell(cmd.Cmd):
    def do_bad(self, arg):
        raise Exception('something bad happened')


if __name__ == '__main__':
    sh = myShell()
    sh.cmdloop()

，我想返回引发异常后转到shell提示符。

and I want to return to the shell-prompt after the exception was thrown. How to do that?

推荐答案

从代码中，从Cmd.onecmd（甚至在循环中）调用函数。

From the code, the functions are called from Cmd.onecmd (even in loop).

您可以简单地覆盖它：

def onecmd(self, line):
    try:
        return super().onecmd(line)
    except:
        # display error message
        return False # don't stop

优点是您不必停止命令循环。

The advantage is that you don't stop the command loop.

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