在bash中,请参阅help getopts
:“当选项需要参数时,getopts将该参数放入shell变量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,你可以围绕谷歌的解决方案来解析在bash选项。这里有一个几分钟的努力:
#!/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 $port"
echo "the rest of the args are:"; (IFS=,; echo "$*")
和测试,
$ 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
当然,你实际上调用命令'的MyScript -p 1984年-n someName' –