How do I measure duration in seconds in a shell script?

Question

I wish to find out how long an operation takes in a Linux shell script. How can I do this?

Answer 1

Many of the answers mention $SECONDS, but that variable is actually even better than they realize:

Assignment to this variable resets the count to the value assigned, and the expanded value becomes the value assigned plus the number of seconds since the assignment.

This means you can simply query this variable directly at the end of your script to print the elapsed time:

#!/usr/bin/env bash

# Do stuff...

echo "Script finished in $SECONDS seconds."

You can also time smaller sections like so:

#!/usr/bin/env bash

# Do stuff

SECONDS=0

# Do timed stuff...

echo "Timed stuff finished in $SECONDS seconds."

Answer 2

Using the time command, as others have suggested, is a good idea.

Another option is to use the magic built-in variable $SECONDS, which contains the number of seconds since the script started executing. You can say:

START_TIME=$SECONDS
dosomething
ELAPSED_TIME=$(($SECONDS - $START_TIME))

I think this is bash-specific, but since you're on Linux, I assume you're using bash.

Answer 3

Use the time command. time ls /bin.

Answer 4

Here is the script to find the time elapsed in milliseconds. Replace the sleep 60 line with the code you want to execute.

a=0
while [ $a -lt 10 ]
do
START_TIME=`echo $(($(date +%s%N)/1000000))`
sleep 3
END_TIME=`echo $(($(date +%s%N)/1000000))`
ELAPSED_TIME=$(($END_TIME - $START_TIME))
echo $ELAPSED_TIME
if [ $a -eq 10 ]
then
  break
fi
a=`expr $a + 1`
done

Answer 5

Try following example:

START_TIME=$SECONDS
# do something
sleep 65

ELAPSED_TIME=$(($SECONDS - $START_TIME))

echo "$(($ELAPSED_TIME/60)) min $(($ELAPSED_TIME%60)) sec"    
#> 1 min 5 sec

Answer 6

Just to help anyone like me that receive an error:

 arithmetic expression: expecting primary: "-"

Check your shellscript that shall start with:

#!/bin/bash

Cheers!