[英]How to get the total execution time of a previously executed shell script in Linux?
[英]Calculate time for each step of a shell script and show total execution time
我有下面的脚本,并且要求我必须为每个脚本放置一些函数来获取每个脚本的时间信息,并最后显示总时间。
我的主要脚本如下所示:
/u01/scripts/stop.sh ${1} | tee ${stop_log}
/u01/scripts/kill_proc.sh ${1} | tee ${kill_log}
/u01/scripts/detach.sh ${1}| tee ${detach_log}
/u01/scripts/copy.sh ${1} | tee ${copy_log}
我想使用类似下面的函数来获取每个脚本执行时间,最后使用全局变量我可以显示所有脚本花费的总时间。
我在下面创建但不幸的是我无法正常使用,如果你在这里有一些善意的帮助。
time_check()
{
export time_log=${log}/time_log_${dts}.log
echo 'StartingTime:'date +%s > ${time_log}
echo 'EndingTime:'date +%s >> ${time_log}
}
我想使用类似于上面的函数来获取每个脚本执行时间,最后使用全局变量我可以显示所有脚本所花费的总时间。 任何人都可以指导如何获得所需的结果。
如果您对时间粒度为秒的确定,则可以执行以下操作:
start=$SECONDS
/u01/scripts/stop.sh ${1} | tee ${stop_log}
stop=$SECONDS
/u01/scripts/kill_proc.sh ${1} | tee ${kill_log}
kill_proc=$SECONDS
/u01/scripts/detach.sh ${1}| tee ${detach_log}
detach=$SECONDS
/u01/scripts/copy.sh ${1} | tee ${copy_log}
end=$SECONDS
printf "%s\n" "stop=$((stop-start)), kill_proc=$((kill_proc-stop)), detach=$((detach-kill_proc)), copy=$((end-detach)), total=$((end-start))"
您也可以编写一个函数来执行此操作:
time_it() {
local start=$SECONDS rc
echo "$(date): Starting $*"
"$@"; rc=$?
echo "$(date): Finished $*; elapsed = $((SECONDS-start)) seconds"
return $rc
}
使用Bash版本> = 4.2,您可以使用printf
打印日期而不是调用外部命令:
time_it() {
local start=$SECONDS ts rc
printf -v ts '%(%Y-%m-%d_%H:%M:%S)T' -1
printf '%s\n' "$ts Starting $*"
"$@"; rc=$?
printf -v ts '%(%Y-%m-%d_%H:%M:%S)T' -1
printf '%s\n' "$ts Finished $*; elapsed = $((SECONDS-start)) seconds"
return $rc
}
并将其调用为:
start=$SECONDS
time_it /u01/scripts/stop.sh ${1} | tee ${stop_log}
time_it /u01/scripts/kill_proc.sh ${1} | tee ${kill_log}
time_it /u01/scripts/detach.sh ${1}| tee ${detach_log}
time_it /u01/scripts/copy.sh ${1} | tee ${copy_log}
echo "Total time = $((SECONDS-start)) seconds"
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