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Top Forums Shell Programming and Scripting Getting the Start, End time and duration using date command Post 302724891 by omega3 on Thursday 1st of November 2012 12:06:49 PM
Old 11-01-2012
Thanks Bipinajith. I tested your script with sleep for 2 minutes. The results are not quite what I wanted. I didn't get the minutes part right.
But I'll work on it. Thank you.
Code:
 
$ echo $0
-ksh
$ cat test.sh
START_DATE=`date`
sleep 120
END_DATE=`date`
 
echo "Job started at: $START_DATE"
echo "Job started at: $END_DATE"
sdt=$( date -d "$START_DATE" +%s )
edt=$( date -d "$END_DATE" +%s )
echo "$edt $sdt" | awk ' { printf("It took %.2f hours to complete this job.\n", ($1-$2)/60/60); } '
$ ./test.sh
Job started at: Thu Nov  1 23:53:38 HKT 2012
Job started at: Thu Nov  1 23:55:38 HKT 2012
It took 0.03 hours to complete this job.

 

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DATETIME.SUB(3) 							 1							   DATETIME.SUB(3)

DateTime::sub - Subtracts an amount of days, months, years, hours, minutes and seconds from a DateTime object

       Object oriented style

SYNOPSIS
public DateTime DateTime::sub (DateInterval $interval) DESCRIPTION
Procedural style DateTime date_sub (DateTime $object, DateInterval $interval) Subtracts the specified DateInterval object from the specified DateTime object. PARAMETERS
o $object -Procedural style only: A DateTime object returned by date_create(3). The function modifies this object. o $interval - A DateInterval object RETURN VALUES
Returns the DateTime object for method chaining or FALSE on failure. EXAMPLES
Example #1 DateTime.sub(3) example Object oriented style <?php $date = new DateTime('2000-01-20'); $date->sub(new DateInterval('P10D')); echo $date->format('Y-m-d') . " "; ?> Procedural style <?php $date = date_create('2000-01-20'); date_sub($date, date_interval_create_from_date_string('10 days')); echo date_format($date, 'Y-m-d'); ?> The above examples will output: 2000-01-10 Example #2 Further DateTime.sub(3) examples <?php $date = new DateTime('2000-01-20'); $date->sub(new DateInterval('PT10H30S')); echo $date->format('Y-m-d H:i:s') . " "; $date = new DateTime('2000-01-20'); $date->sub(new DateInterval('P7Y5M4DT4H3M2S')); echo $date->format('Y-m-d H:i:s') . " "; ?> The above example will output: 2000-01-19 13:59:30 1992-08-15 19:56:58 Example #3 Beware when subtracting months <?php $date = new DateTime('2001-04-30'); $interval = new DateInterval('P1M'); $date->sub($interval); echo $date->format('Y-m-d') . " "; $date->sub($interval); echo $date->format('Y-m-d') . " "; ?> The above example will output: 2001-03-30 2001-03-02 NOTES
DateTime.modify(3) is an alternative when using PHP 5.2. SEE ALSO
DateTime.add(3), DateTime.diff(3), DateTime.modify(3). PHP Documentation Group DATETIME.SUB(3)
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