Prevent script from outputitng status code - bash

Question

I wrote some ugly script of mine and besides giving me what I want in the output it also gives me the status code value.
The script output is below. How do I prevent the script from showing that status code in the output??
P.S. I'll put the script below, hope it wouldn't hurt your eyes too much
Status code = 0
Status code = 0
Status code = 0
job_1_1_1
Status code = 0
Status code = 0
Status code = 0
Status code = 0
Status code = 0
Status code = 0
job_1_1_2
Status code = 0
Status code = 0
START_ID=`dsjob -logsum -type STARTED UPSTREAM_MDM_D4 seq_1_1 | nawk 'ORS=(FNR%2)?FS:RS' | grep Starting | tail -1 | awk '{print $1 }'`; FATAL_IDS=`dsjob -logsum -type INFO UPSTREAM_MDM_D4 seq_1_1 | grep INFO | awk '{print $1 }'`; for TEST_ID in ${FATAL_IDS}; do if [[ "${TEST_ID}" -ge "${START_ID}" ]]; then WARN_DTL=`dsjob -logdetail UPSTREAM_MDM_D4 seq_1_1 ${TEST_ID}`; if `echo $WARN_DTL|grep -q 'has finished, status = 3'`; then message=`echo $WARN_DTL| grep -oP '\w+(?= has finished, status = 3)'`; fatal_errors=$fatal_errors$'\n'$message;fi; fi;done; echo $fatal_errors

Answer 1

You should decide what you want exactly achieve, one of method can be for example:
(
your commands
) 2>&1 | grep -v 'Status code = 0'
Beware - the above is not a good practice - only a quick-hack like solution...