synchronizing access 2 variables in a loop in bash

Question

peterc@peterc-ml video $ VAR1="foo maltese mountain"
peterc@peterc-ml video $ VAR2="bar falcon goat"
peterc@peterc-ml video $ for X in 1 2 3
> do
> echo ??????
> done

The desired output is:

 foo bar
 maltese falcon
 mountain goat

Both VAR1 and VAR2 will always have the same number of elements.

Kevin · Accepted Answer · 2012-02-03 19:23:29Z

2

Make an array out of each.

ARR1=( $VAR1 )
ARR2=( $VAR2 )
for (( i = 0; i < ${#ARR1[@]}; i++)); do 
    echo ${ARR1[i]} ${ARR2[i]}
done

e.g.

$ VAR1="a b c"
$ VAR2="1 2 3"
$ ARR1=( $VAR1 )
$ ARR2=( $VAR2 )
$ for (( i = 0; i < ${#ARR1[@]}; i++)); do 
>     echo ${ARR1[i]} ${ARR2[i]}
> done
a 1
b 2
c 3

answered Feb 3, 2012 at 19:23

Kevin

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Carl Norum · Accepted Answer · 2012-02-03 19:19:59Z

1

You can use cut(1):

echo ${VAR1} | cut -d " " -f ${X}

answered Feb 3, 2012 at 19:19

Carl Norum

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2 Comments

Carl Norum Over a year ago

Drive-by downvote? This example does work, so it's hardly wrong.

gavenkoa Over a year ago

Nice to see downvoting of right solution )) Nobody want to think how to get complete solution from this one... +1

glenn jackman · Accepted Answer · 2012-02-03 20:23:02Z

0

paste -d ' ' <(tr ' ' '\n' <<< "$VAR1") <(tr ' ' '\n' <<< "$VAR2")

answered Feb 3, 2012 at 20:23

glenn jackman

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