Bash loop to fetch different variables

Question

Is it possible to do something like this

#!/bin/bash 
noOfParameters=2
paramname1="test1"
paramname2="test2"
i=0
while [ $i -ne ${noOfParameters} ]
  do 
     i=`expr $i + 1`
     echo ${paramname$i}
 done

I am trying to achieve output as

test1
test2

I am getting "main.sh: line 10: ${paramname$i}: bad substitution" error

This looks like a duplicate of "Assigning dynamic bash variable names using a for loop seq" (solution: use an array). Also, use bash's builtin math, like i=$((i + 1)) or just ((i++)), instead of expr. — Gordon Davisson
– Gordon Davisson, Commented Jul 22, 2020 at 21:47

Gilles Quénot · Accepted Answer · 2020-07-22 22:28:12Z

1

I'd personally do this using an array

arr[0]="test"
arr[1]="test2"

for item in "${arr[@]}"
do
  printf "%s\n" "$item"
done

edited Jul 22, 2020 at 22:28

Gilles Quénot

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answered Jul 22, 2020 at 22:05

Lewis Farnworth

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Comments

Walter A · Accepted Answer · 2020-07-27 22:03:05Z

0

You can use indirect references like

for ((i=1;i<3;i++)); do
   varname=paramname$i
   echo "${!varname}"
done

Perhaps you can bypass your problems with something like

set | grep -E "paramname[12]=" | cut -d"=" -f2

answered Jul 27, 2020 at 22:03

Walter A

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