命名 bash 函数参数

Question

一直想知道这一点，但从未充分调查过 - 有没有办法在 bash 中获取命名参数？

例如，我有这个：

function ql_maybe_fail {
  if [[ "$1" == "true" ]]; then
      echo "quicklock: exiting with 1 since fail flag was set for your 'ql_release_lock' command. "
      exit 1;
  fi
}

是否有可能将其转换为这样的东西：

function ql_maybe_fail (isFail) {
  if [[ "$isFail" == "true" ]]; then
      echo "quicklock: exiting with 1 since fail flag was set for your 'ql_release_lock' command. "
      exit 1;
  fi
}

Answer 1

Bash 中的函数当前不支持用户命名参数。

Answer 2

这个解决方法可能会有所帮助，但它不是很好的测试：

fun () {
    v1=$1
    v2=$2
    for v in "$v1" "$v2"
    do
       case "$v" in
           name=*) name=${v/*=/};;
           age=*)  age=${v/*=/};;
           *)    echo "unexpected $v, please use name and age" ;;
       esac
    done

    echo "name=$name age=$age"
}

输出：

fun "name=John" "age=22"
name=John age=22
fun "age=22" "name=John"
name=John age=22