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?

Solution 1

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]=

Solution 2

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

Comments

[VextoR]

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

[UmNyobe]

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