bash: find string in multiple other strings

Question

I'm trying to use bash to find if a short string is present in any string "sets". For example,

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

MY_CHOICE="kiwi"
MY_CHOICE_GROUP="?"

How can I set MY_CHOICE_GROUP to FRUIT?

I tried to use this StackOverflow solution, but it only works with a single string set.

Originally, I was using arrays to store the options in a set, but given the way bash handles iteration over arrays, it seems a string search would be more efficient.

Many thanks!

Are the sets 'given' (unchanging)? – sehe Jul 17 '13 at 22:23 — sehe, Jul 17 '13 at 22:23
Yes, they are given (they won't change). – hao_maike Jul 17 '13 at 22:24 — hao_maike, Jul 17 '13 at 22:24

sehe · Accepted Answer · 2013-07-17T22:40:24.340

The simplest way, IMO, would be to just hardcode a bunch of case...esac labels.

#!/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

See it live on ideone

Otherwise, consider associative arrays, see it 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

+1 for two solutions, though would be nice to mention that second one is `bash` version 4 or later only. — jaypal singh, Jul 18 '13 at 01:15

score 1 · Answer 2 · answered Jul 18 '13 at 00:14

Only shortening a bit @sehe's answer:

#!/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

pmoosh · Answer 3 · 2013-07-18T19:38:46.490

1

#!/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

edited Jul 18 '13 at 19:38

answered Jul 18 '13 at 00:43

pmoosh

60
5

+1 for using variable indirection, but finish the job in bash: `if [[ ${!group} == *${MY_CHOICE}* ]]; ...` – glenn jackman Jul 18 '13 at 03:21

bash: find string in multiple other strings

3 Answers3