Select random variable in Bash

Question

I need print value of (i.e. CadetBlue) random var from COLOUR_* I tried"

echo $COLOUR_${NUMBER}
echo "$COLOUR_${NUMBER}"

and many others and none is working.

I Have bash:

#!/bin/bash
NUMBER=$[ ( $RANDOM % 9 )  + 1 ]
echo $NUMBER

COLOUR_1=AliceBlue
COLOUR_2=AntiqueWhite
COLOUR_3=AntiqueWhite1
COLOUR_4=AntiqueWhite2
COLOUR_5=AntiqueWhite3
COLOUR_6=AntiqueWhite4
COLOUR_7=BlanchedAlmond
COLOUR_8=BlueViolet
COLOUR_9=CadetBlue

echo $COLOUR_$NUMBER

$[...] is a deprecated syntax; use $((...)) instead.

chepner
– chepner

2013-07-18 12:30:11 +00:00
Commented Jul 18, 2013 at 12:30 — chepner
– chepner, Commented Jul 18, 2013 at 12:30

devnull · Accepted Answer · 2013-07-18 11:16:58Z

5

Instead of saying

echo $COLOUR_$NUMBER

say:

color=COLOUR_${NUMBER}
echo ${!color}

You can read more about indirect expansion here.

answered Jul 18, 2013 at 11:16

devnull

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Comments

chepner · Accepted Answer · 2013-07-18 12:29:17Z

5

Or use an array:

#!/bin/bash
NUMBER=$(( $RANDOM % 9 ))
echo $NUMBER

COLOURS=(AliceBlue AntiqueWhite AntiqueWhite1 AntiqueWhite2
    AntiqueWhite3 AntiqueWhite4 BlanchedAlmond BlueViolet CadetBlue)

echo ${COLOURS[$NUMBER]}

edited Jul 18, 2013 at 12:29

chepner

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answered Jul 18, 2013 at 11:16

perreal

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1 Comment

glenn jackman Over a year ago

+1 although you don't need the line continuation when declaring the array.

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