shell命令返回值判断
ghostwritten 人气:01.判断命令是否存在
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。）
优雅方法2
将通用逻辑转移到共享函数中。
check() {
local command=("$@")
if "${command[@]}"; then
echo notify user OK >&2
else
echo notify user FAIL >&2
exit 1
fi
}
check command1
check command2
check command3优雅方法3
installed () {
command -v "$1" >/dev/null 2>&1
}
if installed <command1>
then
<command1> xx
else
<command1> xxx
fi2.返回错误退出
1.|| exit退出
command1 || exit command2 || exit command3 || exit
2.使用-e
$ bash -e xx.sh #!/bin/bash -e xx.sh command1 command2 command3
3.set -e
$ bash xx.sh #!/bin/bash set -e command1 command2 command3
3.返回错误提示
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
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
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