从输出被捕获的函数中退出 bash 脚本

Question

我想在出错时完全终止/退出 bash shell 脚本，但使用函数error 让我在终止前显示调试输出。现在我遇到的问题是，如果函数的输出是通过反引号或$() 捕获的，则错误函数中的exit 1 语句将不终止shell 脚本。

这是我的示例脚本：

    #!/bin/bash

    function error ()
    {
            echo "An error has occured: $1"
            exit 1
    }
    function do_sth ()
    {
            if [ $1 -eq 0 ]; then
                    error "First param must be greater than 0!"
            else
                    echo "OK!"
            fi
    }

    RESULT=`do_sth 0`

    echo "This line should never be printed"

如何立即终止 error() 函数中的脚本？

Answer 1

command substitution 的问题是，一个子shell 开始执行do_sth。 exit 1 然后终止这个 subshel​​l 而不是主 bash。

您可以通过附加|| exit $? 来解决此问题，它使用命令替换的退出代码退出

RESULT=`do_sth 0` || exit $?

如果要显示错误信息，请将其重定向到 stderr

echo "An error has occured: $1" >&2

Answer 2

RESULT=`do_sth 0` || exit $?

那么 echo "发生错误：$1" >&2

Answer 3

没有办法避免子shell 不能直接终止父进程的事实：父进程必须评估从子shell 返回的值。一种常见的技术是陷阱退出并在陷阱函数中打印错误消息。由于您在子 shell 中生成错误消息，因此您不能像其他方式那样简单地将消息分配给变量，但您可以使用文件系统。请注意，这个想法真的很傻，简单地将错误消息写入stderr要干净得多。这就是它的用途，这就是它被孩子继承的原因。比如：

#!/bin/sh

trap final 0 2 15

# Create a temporary file to store error messages.  This is a terrible
# idea: it would be much better to simply write error messages to stderr,
# but this code is attempting to demonstrate the technique of having the
# parent print the message.  Perhaps it would do better to serve as an example
# of why reporting your children's mistakes is a bad idea.  The children
# should be responsible for reporting their own errors.  Doing so is easy,
# since they inherit file descriptor 2 from their parent.
errmsg=$( mktemp xxxxx )
final() {
    test "$?" = 0 || cat $errmsg
    rm -f $errmsg
} >&2

# Must emphasize one more time that this is a silly idea.  The error 
# function ought to be writing to stderr: eg echo "error: $*" >&2
error() { echo "error: $*" > $errmsg; exit 1; }
do_sth() { 
    if test "$1" -eq 0; then
        error "First param must be greater than 0!"
    else
        echo "OK!"
    fi
}


result=$( do_sth 0 ) || exit 1
echo not printed

Answer 4

似乎（聪明的）答案是@FatalError，here on so

的另一个问题的最佳答案

Answer 5

这应该可以...

#!/bin/bash

trap "exit 1" 50                                       #exit process after receiving signal 50.

function myerror ()
{
    echo "An error has occured: $1" >&2
}
function do_sth ()
{
    if [ $1 -eq 0 ]; then
        myerror "First param must be greater than 0!"
        kill -50 $(ps --pid $$ -opid=)                 #uncommon signal 50 is used.
    else
        echo "OK!"
    fi
}

RESULT=`do_sth 1`
echo $RESULT

RESULT=`do_sth 0`
echo $RESULT

echo "This line should never be printed"