[SHELL]判断一个命令是否存在
首先要说明的是,不要使用which来进行判断,理由如下:
1、which非SHELL的内置命令,用起来比内置命令的开销大,并且非内置命令会依赖平台的实现,不同平台的实现可能不同。
# type type type is a shell builtin # type command command is a shell builtin # type which which is hashed (/usr/bin/which)
2、很多系统的which并不设置退出时的返回值,即使要查找的命令不存在,which也返回0
# which ls /usr/bin/ls # echo $? 0 # which aaa no aaa in /usr/bin /bin /usr/sbin /sbin /usr/local/bin /usr/local/bin /usr/local/sbin /usr/ccs/bin /usr/openwin/bin /usr/dt/bin # echo $? 0
3、许多系统的which实现,都偷偷摸摸干了一些“不足为外人道也”的事情
所以,不要用which,可以使用下面的方法:
$ command -v foo >/dev/null 2>&1 || { echo >&2 "I require foo but it's not installed. Aborting."; exit 1; } $ type foo >/dev/null 2>&1 || { echo >&2 "I require foo but it's not installed. Aborting."; exit 1; } $ hash foo 2>/dev/null || { echo >&2 "I require foo but it's not installed. Aborting."; exit 1; }
犀利的原文,可以在这里查看:
http://*.com/questions/592620/how-to-check-if-a-program-exists-from-a-bash-script/677212#677212