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'

通过保持while循环，proc每次都在早上9点由cron启动。但是星期天，我必须在14:00睡觉时保持3小时。你能告诉我如何让它在周日睡3小时。感谢您的帮助！

Answer 1

您可以尝试创建另一个脚本（如监视器脚本）来执行您发布的脚本： monitor_script：将它放在crontab中并每分钟执行一次 procedure_script：你的脚本

monitor_script的目的是：

检查您的procedure_script是否已经＆＃34; up＆amp;运行＆＃34;
如果出于某种原因将执行procedure_script
在monitor_script中插入＆＃34;如果＆＃34;喜欢：如果[$ DAYOFWEEK -eq 6]＆amp;＆amp; [$ yourTimeVar ==＆＃34; 14＆＃34; ];然后睡了3个小时网络

此致 ç

unix脚本应该在工作日（星期一）和周日（9-14和17-21小时）运行9-21小时。它在oracle中执行一个程序

1 个答案: