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Top Forums Shell Programming and Scripting current date - 8 months in perl script Post 302694323 by kmanivan82 on Thursday 30th of August 2012 11:30:45 AM
Old 08-30-2012
current date - 8 months in perl script

I have a requirement as follows.

when i pass a date to the perl script, it has to calculate the current date - 8 months and output the date back to the shell script in date format (YYYY-MM-DD).

Current date - 8 months is not constant.. because leap year, and the months jan, mar, may,.... has 31 days and april, june,.... has 30 days.

The output should be accurate.

Can anyone provide a solution for this issue.

Thanks
Krishnakanth
 

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DATETIME.ADD(3) 							 1							   DATETIME.ADD(3)

DateTime::add - Adds an amount of days, months, years, hours, minutes and seconds to a DateTime object

       Object oriented style

SYNOPSIS
public DateTime DateTime::add (DateInterval $interval) DESCRIPTION
Procedural style DateTime date_add (DateTime $object, DateInterval $interval) Adds the specified DateInterval object to 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.add(3) example Object oriented style <?php $date = new DateTime('2000-01-01'); $date->add(new DateInterval('P10D')); echo $date->format('Y-m-d') . " "; ?> Procedural style <?php $date = date_create('2000-01-01'); date_add($date, date_interval_create_from_date_string('10 days')); echo date_format($date, 'Y-m-d'); ?> The above examples will output: 2000-01-11 Example #2 Further DateTime.add(3) examples <?php $date = new DateTime('2000-01-01'); $date->add(new DateInterval('PT10H30S')); echo $date->format('Y-m-d H:i:s') . " "; $date = new DateTime('2000-01-01'); $date->add(new DateInterval('P7Y5M4DT4H3M2S')); echo $date->format('Y-m-d H:i:s') . " "; ?> The above example will output: 2000-01-01 10:00:30 2007-06-05 04:03:02 Example #3 Beware when adding months <?php $date = new DateTime('2000-12-31'); $interval = new DateInterval('P1M'); $date->add($interval); echo $date->format('Y-m-d') . " "; $date->add($interval); echo $date->format('Y-m-d') . " "; ?> The above example will output: 2001-01-31 2001-03-03 NOTES
DateTime.modify(3) is an alternative when using PHP 5.2. SEE ALSO
DateTime.sub(3), DateTime.diff(3), DateTime.modify(3). PHP Documentation Group DATETIME.ADD(3)
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