bash - 如何在 bash 中使用 getopts 的示例

Question

我想这样调用myscript文件:

$ ./myscript -s 45 -p any_string

或

$ ./myscript -h  #should display help
$ ./myscript     #should display help

我的要求是:

• getopt 获取输入参数
• 检查-s是否存在，如果不存在则返回错误
• 检查 -s 之后的值是 45 还是 90
• 检查-p是否存在并且后面有输入字符串
• 如果用户输入 ./myscript -h 或只是 ./myscript 然后显示帮助

到目前为止我试过这段代码:

#!/bin/bash
while getopts "h:s:" arg; do
  case $arg in
    h)
      echo "usage" 
      ;;
    s)
      strength=$OPTARG
      echo $strength
      ;;
  esac
done

但是使用该代码我会出错。如何使用 Bash 和 getopt 做到这一点？

Answer 1

#!/bin/bash

usage() { echo "Usage: $0 [-s <45|90>] [-p <string>]" 1>&2; exit 1; }

while getopts ":s:p:" o; do
    case "${o}" in
        s)
            s=${OPTARG}
            ((s == 45 || s == 90)) || usage
            ;;
        p)
            p=${OPTARG}
            ;;
        *)
            usage
            ;;
    esac
done
shift $((OPTIND-1))

if [ -z "${s}" ] || [ -z "${p}" ]; then
    usage
fi

echo "s = ${s}"
echo "p = ${p}"

运行示例:

$ ./myscript.sh
Usage: ./myscript.sh [-s <45|90>] [-p <string>]

$ ./myscript.sh -h
Usage: ./myscript.sh [-s <45|90>] [-p <string>]

$ ./myscript.sh -s "" -p ""
Usage: ./myscript.sh [-s <45|90>] [-p <string>]

$ ./myscript.sh -s 10 -p foo
Usage: ./myscript.sh [-s <45|90>] [-p <string>]

$ ./myscript.sh -s 45 -p foo
s = 45
p = foo

$ ./myscript.sh -s 90 -p bar
s = 90
p = bar