在bash中,請參閱help getopts
:「當選項需要參數時,getopts將該參數放入shell變量OPTARG。」
usage() { echo "Usage: $(basename $0) -n name -p port -r"; exit; }
while getopts :n:p:r opt # don't forget the colons for opts that take an arg
do
case $opt in
n) name="$OPTARG" ;;
p) port="$OPTARG" ;;
r) robot=chicken ;;
?) usage ;;
esac
done
shift $((OPTIND - 1))
echo "the name is $name"
echo "the port is $port"
我敢肯定,你可以圍繞谷歌的解決方案來解析在bash選項。這裏有一個幾分鐘的努力:
#!/bin/bash
usage() { echo foo; exit; }
while [[ $1 == -* ]]; do
case "$1" in
--) shift 1; break ;;
-p|--p|--port) port="$2"; shift 2;;
-n|--n|--name) name="$2"; shift 2;;
*) echo "unknown option: $1"; usage;;
esac
done
echo "the name is $name"
echo "the port is $port"
echo "the rest of the args are:"; (IFS=,; echo "$*")
和測試,
$ bash longopts.sh --port 1234 --bar a b c
unknown option: --bar
foo
$ bash longopts.sh --port 1234 a b c
the name is
the port is 1234
the rest of the args are:
a,b,c
當然,你實際上調用命令'的MyScript -p 1984年-n someName' –