从捕获输出的函数中退出bash脚本

Question

我想在发生错误时完全终止/退出bash shell脚本，但使用的函数error使我可以在终止前显示调试输出。 现在，我遇到的问题是，如果通过反引号或$()捕获函数的输出，则错误函数内的exit 1语句将不会终止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"

如何立即在error()函数中终止脚本？

Answer 1

命令替换的问题是，启动了一个子do_sth来执行do_sth 。 然后， exit 1终止此子shell，而不终止主bash。

您可以通过添加|| exit $?解决此问题。 || exit $? ，它以命令替换中的退出代码退出

RESULT=`do_sth 0` || exit $?

如果要显示错误消息，请将其重定向到stderr

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

Answer 2

RESULT=`do_sth 0` || exit $?

然后回显“发生错误：$ 1”>＆2

Answer 3

无法避免子shell无法使父对象直接终止的事实：父对象必须评估从子shell返回的值。 一种常见的技术是在退出功能中捕获退出并打印错误消息。 由于您是在子外壳程序中生成错误消息，因此不能像其他情况那样简单地将消息分配给变量，而是可以使用文件系统。 请注意，这种想法确实很愚蠢，仅将错误消息写入stderr会更清洁。 这就是它的用途，这就是为什么它被孩子们继承的原因。 就像是：

#!/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）答案是@FatalError在另一个问题的最高答案上，因此

Answer 5

这应该工作...

#!/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"