CODEWITHSUNDEEP
linuxbashshellunixscripting
user2500742
user2500742

Reputation: 811

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.

My main scripts looks like below: 

/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.

I created below but unfortunately I could not use properly , if you have something kindly help here.

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 . Could anyone please guide how to get the desired result.

Upvotes: 5

Views: 1542

Answers (1)

codeforester
codeforester

Reputation: 42979

If you are OK with the time granularity of seconds, you could simply do this:

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:

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:

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:

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:

Upvotes: 2

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