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
– sehe

2013-07-17 22:23:52 +00:00
Commented Jul 17, 2013 at 22:23 — sehe
– sehe, Commented Jul 17, 2013 at 22:23
Yes, they are given (they won't change).

hao_maike
– hao_maike

2013-07-17 22:24:28 +00:00
Commented Jul 17, 2013 at 22:24 — hao_maike
– hao_maike, Commented Jul 17, 2013 at 22:24

sehe · Accepted Answer · 2013-07-17 22:40:24Z

6

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

edited Jul 17, 2013 at 22:40

answered Jul 17, 2013 at 22:29

sehe

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1 Comment

jaypal singh Over a year ago

+1 for two solutions, though would be nice to mention that second one is bash version 4 or later only.

clt60 · Accepted Answer · 2013-07-18 00:14:38Z

1

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

answered Jul 18, 2013 at 0:14

clt60

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Comments

pmoosh · Accepted Answer · 2013-07-18 19:38:46Z

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, 2013 at 19:38

answered Jul 18, 2013 at 0:43

pmoosh

605 bronze badges

1 Comment

glenn jackman Over a year ago

+1 for using variable indirection, but finish the job in bash: if [[ ${!group} == *${MY_CHOICE}* ]]; ...

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