date_sub

(No version information available, might be only in CVS)

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

Description

void date_sub ( DateTime $object , DateInterval $interval )

Subtracts the specified DateInterval object from the specified DateTime object.

Parameters

object

A date as returned by DateTime.

interval

The amount to be subtracted. For the date use "P3D", "P3M", "P3Y" or a combination of the three e.g. "P2M5D" (Y = Years, M = Months, D = Days.) MUST BE YEAR MONTH DAY FORMAT "P5Y", "P5M2D", "P5Y4D". For the time use "T3H", "T3M", "T3S" or or a combination of the three e.g. "T5H20M" (H = Hours, M = Minutes, S = Seconds). For dateTime us "P5D2M4YT5H20M". The digit before the letter (NOT P or T) can be any amount.

Return Values

No value is returned.

Examples

Example #1 date_sub() example

<?php

$date 
= new DateTime("18-July-2008 16:30:30");
echo 
$date->format("d-m-Y H:i:s").'<br />';

date_sub($date, new DateInterval("P5D"));
echo 
'<br />'.$date->format("d-m-Y").' : 5 Days';

date_sub($date, new DateInterval("P5M"));
echo 
'<br />'.$date->format("d-m-Y").' : 5 Months';

date_sub($date, new DateInterval("P5Y"));
echo 
'<br />'.$date->format("d-m-Y").' : 5 Years';

date_sub($date, new DateInterval("P5Y5M5D"));
echo 
'<br />'.$date->format("d-m-Y").' : 5 Days, 5 Months, 5 Years';

date_sub($date, new DateInterval("P5YT5H"));
echo 
'<br />'.$date->format("d-m-Y H:i:s").' : 5 Years, 5 Hours';

?>

Notes

Warning

This function is EXPERIMENTAL. The behaviour of this function, its name, and surrounding documentation may change without notice in a future release of PHP. This function should be used at your own risk.

See Also


    

 

 

Home || Web Directroy || Web Designing || Get a Domain Name || Host your web site || Free Blogs || Free Tutorial || Free E-cards || Free GuestBook || Email