Question

我正在尝试使用bash来查找任何字符串“sets”中是否存在短字符串。例如，

FRUIT="apple banana kiwi melon"
VEGETABLE="radish lettuce potato"
COLOR="blue red yellow green brown"

MY_CHOICE="kiwi"
MY_CHOICE_GROUP="?"

如何将MY_CHOICE_GROUP设为FRUIT？

我尝试使用this StackOverflow solution，但它仅适用于单个字符串集。

最初，我使用数组来存储集合中的选项，但是given the way bash handles iteration over arrays，似乎字符串搜索会更有效率。

非常感谢！

Answer 1

IMO最简单的方法就是硬编码一堆case...esac标签。

#!/bin/bash
function lookup()
{
    case "$1" in
        apple|banana|kiwi|melon)
            echo "FRUIT"
            ;;

        radish|lettuce|potato)  
            echo "VEGETABLE"
            ;;

        blue|red|yellow|green|brown)   
            echo "COLOR"
            ;;
    esac
}

MY_CHOICE="kiwi"
MY_CHOICE_GROUP=$(lookup "$MY_CHOICE")

echo $MY_CHOICE: $MY_CHOICE_GROUP

查看 live on ideone

否则，请考虑关联数组，请参阅 live on ideone ：

#!/bin/bash
declare -A groups

groups["apple"]="FRUIT"
groups["banana"]="FRUIT"
groups["kiwi"]="FRUIT"
groups["melon"]="FRUIT"

groups["radish"]="VEGETABLE"
groups["lettuce"]="VEGETABLE"
groups["potato"]="VEGETABLE"

groups["blue"]="COLOR"
groups["red"]="COLOR"
groups["yellow"]="COLOR"
groups["green"]="COLOR"
groups["brown"]="COLOR"

MY_CHOICE="kiwi"
MY_CHOICE_GROUP=${groups[$MY_CHOICE]}

echo $MY_CHOICE: $MY_CHOICE_GROUP

Answer 2

只是缩短了一点@ sehe的答案：

#!/bin/bash

declare -A groups
mkaso() { val="$1"; shift; for key in "$@"; do groups["$key"]="$val"; done; }

mkaso FRUIT apple banana kiwi melon
mkaso VEGETABLE radish lettuce potato
mkaso COLOR blue red yellow green brown
#declare -p groups

MY_CHOICE="kiwi"
MY_CHOICE_GROUP=${groups[$MY_CHOICE]}
echo $MY_CHOICE: $MY_CHOICE_GROUP

Answer 3

#!/bin/bash
FRUIT="apple banana kiwi melon"
VEGETABLE="radish lettuce potato"
COLOR="blue red yellow green brown"

MY_CHOICE="kiwi"

for group in VEGETABLE COLOR FRUIT
do
    if [[ ${!group} == *${MY_CHOICE}* ]]; then
        MY_CHOICE_GROUP=$group
        break
    fi
done

echo $MY_CHOICE_GROUP

bash：在多个其他字符串中查找字符串

3 个答案: