出口不在bash工作 [英] exit doesn't work in bash

问题描述

下面是bash脚本

#!/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

和这是结果...

+ 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!

我看到了命令退出，但它并没有退出退出脚本

I saw the command "exit", but it didn't exit exit the script

为什么退出不工作？

推荐答案

您在呼唤退出从子shell里面，所以这是退出外壳。试试这个：

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?"
}

然而，这将从一个名为任何壳退出被叫方。您可能需要使用收益而不是退出从函数返回。

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

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