在不关闭 shell 的情况下返回退出代码 [英] Return an exit code without closing shell

问题描述

我想从在另一个脚本中调用的 BASH 脚本返回退出代码，但也可以直接调用.大致是这样的:

I'd like to return an exit code from a BASH script that is called within another script, but could also be called directly. It roughly looks like this:

#!/bin/bash
dq2-get $1
if [ $? -ne 0 ]; then
  echo "ERROR: ..."
  # EXIT HERE
fi
# extract, do some stuff
# ...

现在在 EXIT HERE 行中，脚本应该退出并返回退出代码 1.问题是

Now in the line EXIT HERE the script should exit and return exit code 1. The problem is that

• 我不能使用 return，因为当我忘记 source 脚本而不是调用它时，return 将 not 退出，并且脚本的其余部分将被执行并把事情搞砸了.
• 我不能使用 exit，因为这会关闭 shell.
• 我不能使用巧妙的技巧 kill -SIGINT $$，因为这不允许返回退出代码.

• I cannot use return, because when I forget to source the script instead of calling it, return will not exit, and the rest of the script will be executed and mess things up.
• I cannot use exit, because this closes the shell.
• I cannot use the nice trick kill -SIGINT $$, because this doesn't allow to return an exit code.

有没有我忽略的可行替代方案?

Is there any viable alternative that I have overlooked?

推荐答案

你可以使用 x"${BASH_SOURCE[0]}" == x"$0" 来测试脚本是否有源或调用(如果来源为 false，如果调用则为 true)并相应地 return 或 exit.

You can use x"${BASH_SOURCE[0]}" == x"$0" to test if the script was sourced or called (false if sourced, true if called) and return or exit accordingly.

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