你可以在非redhat系统上使用@john-kugelman的解决方案，在他的代码中注释掉这一行:

. /etc/init.d/functions

然后，在最后粘贴下面的代码。完全披露:这只是从Centos 7中直接复制和粘贴上述文件的相关位。

在MacOS和Ubuntu 18.04上测试。

BOOTUP=color
RES_COL=60
MOVE_TO_COL="echo -en \\033[${RES_COL}G"
SETCOLOR_SUCCESS="echo -en \\033[1;32m"
SETCOLOR_FAILURE="echo -en \\033[1;31m"
SETCOLOR_WARNING="echo -en \\033[1;33m"
SETCOLOR_NORMAL="echo -en \\033[0;39m"

echo_success() {
    [ "$BOOTUP" = "color" ] && $MOVE_TO_COL
    echo -n "["
    [ "$BOOTUP" = "color" ] && $SETCOLOR_SUCCESS
    echo -n $"  OK  "
    [ "$BOOTUP" = "color" ] && $SETCOLOR_NORMAL
    echo -n "]"
    echo -ne "\r"
    return 0
}

echo_failure() {
    [ "$BOOTUP" = "color" ] && $MOVE_TO_COL
    echo -n "["
    [ "$BOOTUP" = "color" ] && $SETCOLOR_FAILURE
    echo -n $"FAILED"
    [ "$BOOTUP" = "color" ] && $SETCOLOR_NORMAL
    echo -n "]"
    echo -ne "\r"
    return 1
}

echo_passed() {
    [ "$BOOTUP" = "color" ] && $MOVE_TO_COL
    echo -n "["
    [ "$BOOTUP" = "color" ] && $SETCOLOR_WARNING
    echo -n $"PASSED"
    [ "$BOOTUP" = "color" ] && $SETCOLOR_NORMAL
    echo -n "]"
    echo -ne "\r"
    return 1
}

echo_warning() {
    [ "$BOOTUP" = "color" ] && $MOVE_TO_COL
    echo -n "["
    [ "$BOOTUP" = "color" ] && $SETCOLOR_WARNING
    echo -n $"WARNING"
    [ "$BOOTUP" = "color" ] && $SETCOLOR_NORMAL
    echo -n "]"
    echo -ne "\r"
    return 1
} 

对不起，我不能对第一个答案发表评论 但是你应该使用新实例来执行命令:

#!/bin/bash

function check_exit {
    cmd_output=$($@)
    local status=$?
    echo $status
    if [ $status -ne 0 ]; then
        echo "error with $1" >&2
    fi
    return $status
}

function run_command() {
    exit 1
}

check_exit run_command

对于无意中发现这条线索的鱼壳用户。

假设foo是一个不“返回”(回显)值的函数，但它像往常一样设置退出码。 为了避免在调用函数后检查$status，你可以这样做:

foo; and echo success; or echo failure

如果它太长，不能放在一行上:

foo; and begin
  echo success
end; or begin
  echo failure
end

不要创建runner函数或使用set -e，而是使用trap:

trap 'echo "error"; do_cleanup failed; exit' ERR
trap 'echo "received signal to stop"; do_cleanup interrupted; exit' SIGQUIT SIGTERM SIGINT

do_cleanup () { rm tempfile; echo "$1 $(date)" >> script_log; }

command1
command2
command3

该trap甚至可以访问触发它的命令的行号和命令行。变量为$BASH_LINENO和$BASH_COMMAND。

你说的“退出并回复错误”是什么意思?如果您的意思是希望脚本在任何命令失败时立即终止，那么就这样做

set -e    # DON'T do this.  See commentary below.

在脚本的开头(但注意下面的警告)。不要回显错误消息:让失败的命令处理它。换句话说，如果你这样做:

#!/bin/sh

set -e    # Use caution.  eg, don't do this
command1
command2
command3

而command2失败，同时打印一个错误消息到stderr，那么似乎您已经实现了您想要的。(除非我误解了你想要的!)

因此，您编写的任何命令都必须表现良好:它必须向stderr而不是stdout报告错误(问题中的示例代码将错误输出到stdout)，并且在失败时必须以非零状态退出。

However, I no longer consider this to be a good practice. set -e has changed its semantics with different versions of bash, and although it works fine for a simple script, there are so many edge cases that it is essentially unusable. (Consider things like: set -e; foo() { false; echo should not print; } ; foo && echo ok The semantics here are somewhat reasonable, but if you refactor code into a function that relied on the option setting to terminate early, you can easily get bitten.) IMO it is better to write:

 #!/bin/sh

 command1 || exit
 command2 || exit
 command3 || exit

or

#!/bin/sh

command1 && command2 && command3

您可以编写一个函数来启动和测试该命令。假设command1和command2是已设置为命令的环境变量。

function mytest {
    "$@"
    local status=$?
    if (( status != 0 )); then
        echo "error with $1" >&2
    fi
    return $status
}

mytest "$command1"
mytest "$command2"