在 bash 中处理命名参数

Question

下面是脚本。当我用命令运行这个时

./scriptname --deploy --uglify"

我得到了echo "$deploy"as的结果true。但应该是uglify正确的，因为deploy=$2

#!/bin/bash

deploy=true
uglify=true

while (( $# > 1 )); do 
    case $1 in
    --deploy) deploy="$2" echo "$deploy";;
    --uglify) uglify="$2";;
    *) break;
    esac; shift 2
done

Answer 1

你的意思是这样的：

#!/bin/bash

deploy=false
uglify=false

while (( $# >= 1 )); do 
    case $1 in
    --deploy) deploy=true;;
    --uglify) uglify=true;;
    *) break;
    esac;
    shift
done

echo "deploy: $deploy"
echo "uglify: $uglify"

例子

$ ./scriptname
deploy: false
uglify: false


$ ./scriptname --deploy
deploy: true
uglify: false


$ ./scriptname --uglily
deploy: false
uglify: true

$ ./scriptname --depoly --uglily
deploy: true
uglify: true