[英]Calculate time for each step of a shell script and show total execution time
I have below script and for a requirement I have to place some function for each of these script to get time information for each script and at last show total 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。
My main scripts looks like below: 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 ${copy_log}
I want to use something like below function to get every script execution time and at last with a global variable I can show Total time taken by all the scripts. 我想使用类似下面的函数来获取每个脚本执行时间,最后使用全局变量我可以显示所有脚本花费的总时间。
I created below but unfortunately I could not use properly , if you have something kindly help here. 我在下面创建但不幸的是我无法正常使用,如果你在这里有一些善意的帮助。
time_check()
{
export time_log=${log}/time_log_${dts}.log
echo 'StartingTime:'date +%s > ${time_log}
echo 'EndingTime:'date +%s >> ${time_log}
}
I want to use something like above function to get every script execution time and at last with a global variable I can show total time taken by all the scripts . 我想使用类似于上面的函数来获取每个脚本执行时间,最后使用全局变量我可以显示所有脚本所花费的总时间。 Could anyone please guide how to get the desired result.
任何人都可以指导如何获得所需的结果。
If you are OK with the time granularity of seconds, you could simply do this: 如果您对时间粒度为秒的确定,则可以执行以下操作:
start=$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=$((end-start))"
You can write a function to do this as well: 您也可以编写一个函数来执行此操作:
time_it() {
local start=$SECONDS rc
echo "$(date): Starting $*"
"$@"; rc=$?
echo "$(date): Finished $*; elapsed = $((SECONDS-start)) seconds"
return $rc
}
With Bash version >= 4.2 you can use printf
to print date rather than invoking an external command: 使用Bash版本> = 4.2,您可以使用
printf
打印日期而不是调用外部命令:
time_it() {
local start=$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 $rc
}
And invoke it as: 并将其调用为:
start=$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"
Related: 有关:
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.