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://stackoverflow.com/questions/592620/how-to-check-if-a-program-exists-from-a-bash-script/677212#677212

Q.E.D.