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检查linux命令是否存在的正确方式
source link: https://www.lujun9972.win/blog/2017/03/20/%E6%A3%80%E6%9F%A5linux%E5%91%BD%E4%BB%A4%E6%98%AF%E5%90%A6%E5%AD%98%E5%9C%A8%E7%9A%84%E6%AD%A3%E7%A1%AE%E6%96%B9%E5%BC%8F/index.html
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检查linux命令是否存在的正确方式
之前我一直是在用 which 来判断linux命令是否存在的,但是它并不能用来检查内置命令和函数是否存在.
command_exists() {
which "$@" > /dev/null 2>&1
}
command_exists ls
echo "检查外部命令:" $?
function t ()
{
echo "I am a function"
}
command_exists t
echo "检查函数:" $?
command_exists cd
echo "检查内置命令:" $?
command_exists sldj
echo "检查没有的命令:" $?
检查外部命令: 0 检查函数: 1 检查内置命令: 1 检查没有的命令: 1
直到今天看了下daocloud提供的配置docker加速器脚本(https://get.daocloud.io/daotools/set_mirror.sh).
这个脚本里面也实现了一个 command_exists 函数,不过是使用bash内建的 command 命令来实现的.
command_exists() {
command -v "$@" > /dev/null 2>&1
}
使用 command -v 检查命令时,当命令是已定义的函数,内建命令或者 PATH 中能找到的外部命令时都,返回值都是0,否则会返回1.
command_exists() {
command -v "$@" > /dev/null 2>&1
}
command_exists ls
echo "检查外部命令:" $?
function t ()
{
echo "I am a function"
}
command_exists t
echo "检查函数:" $?
command_exists cd
echo "检查内置命令:" $?
command_exists sldj
echo "检查没有的命令:" $?
检查外部命令: 0 检查函数: 0 检查内置命令: 0 检查没有的命令: 1
这确实比用 which 要更好用.
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