Shell 命令返回值判断

1.判断命令是否存在

1.1 优雅方法1

首先，检查命令是否有效的惯用方法直接在if语句中。

if command; then
    echo notify user OK >&2
else
    echo notify user FAIL >&2
    return -1
fi

（良好做法：使用>&2将消息发送给stderr。）

1.2 优雅方法2

将通用逻辑转移到共享函数中。

check() {
    local command=("$@")

    if ""; then
        echo notify user OK >&2
    else
        echo notify user FAIL >&2
        exit 1
    fi
}

check command1
check command2
check command3

1.3 优雅方法3

installed () {
        command -v "$1" >/dev/null 2>&1
}
if installed <command1>
then
       <command1>  xx
else
        <command1>  xxx
 fi

2. 返回错误退出

2.1 || exit 退出

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

2.2 bash -e

$  bash -e xx.sh
#!/bin/bash -e xx.sh
command1
command2
command3

2.3 set -e

$ bash xx.sh 
#!/bin/bash
set -e 
command1
command2
command3

3. 返回错误提示

3.1 一般方法

方法1

if do some command; then
    echo notify user OK
else
    echo notify user fail
    exit 255  # exit code must be unsigned short
fi

方法2

do some command
if [ $? -eq 0 ]; then
    echo notify user OK
else
    echo notify user FAIL
    return -1
fi

3.2 优雅方法

方法1

die() {
    local message=$1

    echo "$message" >&2
    exit 1
}

command1 || die 'command1 failed'
command2 || die 'command2 failed'
command3 || die 'command3 failed'

方法2（推荐）

warn () {
  echo "$@" >&2
}

die () {
  status="$1"
  shift
  warn "$@"
  exit "$status"
}

do some command && echo notify user OK || die 255 Notify user fail