shell - Unix bash script grep loop counter (for)

Question

I am looping our the a grep result. The result contains 10 lines (every line has different content). So the loop stuff in the loop gets executed 10 times.

I need to get the index, 0-9, in the run so i can do actions based on the index.

ABC=(cat test.log | grep "stuff")

counter=0
for x in $ABC
do
        echo $x      
        ((counter++))
        echo "COUNTER $counter"
done

Currently the counter won't really change.

Output:

51209
120049
148480
1211441
373948
0
0
0
728304
0
COUNTER: 1

question from:https://stackoverflow.com/questions/65624106/unix-bash-script-grep-loop-counter-for

Answer 1

Your shell script contains what should be an obvious syntax error.

ABC=(cat test.log | grep "stuff")

This fails with

-bash: syntax error near unexpected token `|'

There is no need to save the output in a variable if you only want to process one at a time (and obviously no need for the useless cat).

grep "stuff" test.log | nl

gets you numbered lines, though the index will be 1-based, not zero-based.

If you absolutely need zero-based, refactoring to Awk should solve it easily:

awk '/stuff/ { print n++, $0 }' test.log

If you want to loop over this and do something more with this information,

awk '/stuff/ { print n++, $0 }' test.log |
while read -r index output; do
    echo index is "$index"
    echo output is "$output"
done

Because the while loop executes in a subshell the value of index will not be visible outside of the loop. (I guess that's what your real code did with the counter as well. I don't think that part of the code you posted will repro either.)