unix script should run 9-21 hrs on weekdays (mon-sat) and on sunday (9-14 and 17-21hrs). It executes a procecudure in oracle

Question

    START_DATE=`date '+%d%m%Y%H%M%S'`
DAYOFWEEK=$(date +"%u")
START_HOUR=$(date +"%H")


run_dt=`date '+%d%m%Y'`
run_hr=`date '+%H%M'`
echo 'run_hr' $run_hr

echo 'Shell script started'
echo ${START_DATE}
echo ${DAYOFWEEK}
echo ${START_HOUR}

while [ $run_hr -lt 21];
do
if [ $DAYOFWEEK = 1 ] || [ $DAYOFWEEK = 2 ] || [ $DAYOFWEEK = 3 ] || [ $DAYOFWEEK = 4 ] || [ $DAYOFWEEK = 5 ] || [ $DAYOFWEEK = 6 ]; then
echo 'inside while'

sqlplus -s xx/abcd<< END2
set timing on; 
select to_char(sysdate,'dd-mm-yyyyhh24:mi') starttime from dual;
#exec my_proc();
select to_char(sysdate,'dd-mm-yyyyhh24:mi') endtime from dual;
END2

sleep 120
let run_hr=`date '+%H%M'`
echo 'run_hr' $run_hr

elif [ $DAYOFWEEK = 7 ]; then
echo 'inside while on sunday'
sqlplus -s xx/abcd<< END2
set timing on; 
select to_char(sysdate,'dd-mm-yyyyhh24:mi') starttime from dual;
#exec my_proc();
select to_char(sysdate,'dd-mm-yyyyhh24:mi') endtime from dual;
END2

sleep 120
if [ $run_hr -eq 14 ]; then
sleep 10800else
let run_hr=`date '+%H%M'`
echo 'run_hr' $run_hr
echo 'Nothing'
fi
else
echo 'off'
fi
done

echo 'Exiting'

by keeping in the while loop the proc runs every time by initiating at 9am in the morning by a cron. but on sunday, i have to keep it for 3 hrs in sleep at 14:00 hrs. can you please tell me how to keep that to sleep for 3 hrs on sunday. Thanks for the help in advance!

Answer 1

You could try to create another script (like a monitor script)that will execute the one that you posted: monitor_script: put it in crontab and executed it every minute procedure_script: your script

The purpose of monitor_script will be to:

1. check if your procedure_script is already "up & running"
2. execute the procedure_script if for some reason will goes down
3. Insert in monitor_script an "if" like: if [ $DAYOFWEEK -eq 6 ] && [ $yourTimeVar == "14" ];then sleep for 3hrs fi

Regards C