为什么我的“退出"命令不能退出我的Bash脚本?

本文介绍了为什么我的“退出"命令不能退出我的Bash脚本?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有这个Bash脚本:

I have this Bash script:

#!/bin/bash set -x function doSomething() { callee echo "It should not go to here!" } function callee() { ( echo "before" ) && (echo "This is callee" && exit 1 ) echo "why I can see this?" } doSomething

这是结果:

+ set -x + doSomething + callee + echo before before + echo 'This is callee' This is callee + exit 1 + echo 'why I can see this?' why I can see this? + echo 'It should not go to here!' It should not go to here!

我看到命令 exit ，但是它没有退出脚本-为什么 exit 不起作用?

I see the command exit, but it doesn't exit the script – why doesn't exit work?

推荐答案

您正在从子外壳程序内部调用 exit ，因此该外壳程序正在退出.尝试以下方法:

You are calling exit from inside a subshell, so that's the shell that is exiting. Try this instead:

function callee() { ( echo "before" ) && { echo "This is callee" && exit 1; } echo "why I can see this?" }

但是，它将从名为 callee 的任何shell中退出.您可能要使用 return 而不是 exit 从函数中返回.

This, however, will exit from whatever shell called callee. You may want to use return instead of exit to return from the function.