返回退出code，而不关闭壳 [英] Return an exit code without closing shell

问题描述

我想从那个叫另一个脚本中的bash脚本返回退出code，但也可以直接调用。它大致是这样的：

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
# ...

在该行现在退出此处脚本应该退出并返回退出code 1的问题是，

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

• 我不能使用收益，因为当我忘了，而不是源调用它，回报将的不的退出，其余的脚本该脚本将被执行，搞搞。


• 我不能使用退出，因为这将关闭外壳。


• 我无法用好的把戏杀-SIGINT $$ ，因为这不允许返回退出code。

• 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.

有没有可行的替代方案，我忽略了？

Is there any viable alternative that I have overlooked?

推荐答案

您可以使用 X$ {BASH_SOURCE [0]}== X$ 0来测试如果脚本是源或所谓的（假，如果​​采购，真要是称呼）和收益或退出相应

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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