change array value while looping in bash

Question

The code

SourceFolder[0]=""
SourceFolder[1]="inbound2"
SourceFolder[2]="inbound3"

for i in "${!SourceFolder[@]}"
do    
    if [ -z "${SourceFolder[$i]}"]; then
        ${SourceFolder[$i]} = "TEST"
    fi
done

${SourceFolder[$i]} = "TEST" - doesn't work

it says

=: command not found

How to change value in current loop index in an array?

Didn't the proposed solution to your previous question work for you? — devnull
– devnull, Commented Dec 18, 2013 at 9:05
@devnull yes, the solution in that question is working. But in that case if array size will change I have to edit code with that additional looping, which I don't like( that's why I've decided to go in this way — VextoR
– VextoR, Commented Dec 18, 2013 at 9:08

Community · Accepted Answer · 2017-05-23 11:58:52Z

3

Because of the first space = is not interpreted as an assignment. There is a full explanation on So.

Btw ${SourceFolder[$i]} evaluate the array element, which is not what you want to do. For instance for the first one it is the empty string.

Replaces with SourceFolder[$i]=

edited May 23, 2017 at 11:58

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answered Dec 18, 2013 at 9:05

UmNyobe

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2 Comments

VextoR Over a year ago

This helped! btw, it has another error in if [ -z "${SourceFolder[$i]}"]; then it says [: missing ]'` why is that?

UmNyobe Over a year ago

again, watch your spaces. should be [ -z "${SourceFolder[$i]}" ]

m0zgen · Accepted Answer · 2019-02-13 12:39:27Z

1

You must change indexnumber in the your array:

ARRAYNAME[indexnumber]=value

ok, you have array is:

array=(one two three)

you can add count to your script for initialize and change element for array indexnumber, example:

#!/bin/bash

count=0
array=(one two three)

for i in ${array[@]}
do
  echo "$i" 
  array[$count]="$i-indexnumber-is-$count"
  count=$((count + 1))
  echo $count
done

echo ${array[*]}

Result:

bash test-arr.sh 
one
1
two
2
three
3
one-indexnumber-is-0 two-indexnumber-is-1 three-indexnumber-is-2

answered Feb 13, 2019 at 12:39

m0zgen

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