'if' in array w/ Shell Script

Question

I have an array with different values, and if the value is 3 (integer) this value will be exchanged for 999 (integer). But I have a syntax problem in the 'if' block of the array.

The correct return would be: 1 999 5 7 9 999

vetor=(1 3 5 7 9 3)
for i in ${vetor[*]}
do
    if [[ ${vetor[i]} = 3 ]]; then
        ${vetor[i]} = 999
    fi
    echo $i
done

vetor[i]=999 there are no spaces surrounding '=' when used for assignment. You do not use $ (dereference) when making assignment. Also understand [[ ${vetor[i]} = 3 ]] makes a string comparison, while [[ ${vetor[i]} -eq 3 ]] makes a numeric comparison. Changing vetor[i] does not change the loop variable i. — David C. Rankin
– David C. Rankin, Commented May 22, 2021 at 3:09

Nathan Mills · Accepted Answer · 2021-05-22 03:26:54Z

1

This produces the correct output in Bash:

vetor=(1 3 5 7 9 3);
for i in ${!vetor[*]};
do
     if [[ ${vetor[i]} -eq 3 ]]; then
         vetor[i]=999;
     fi;
     echo ${vetor[i]};
done

I added ! in the for loop expression to get the indices of vetor instead of the values, and I removed the ${} around the assignment in the if condition (this was giving "if 3 is not a typo" warnings). Also changed the echo to get the value at vetor[i] instead of printing the index.

answered May 22, 2021 at 3:26

Nathan Mills

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