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HOME > PHP Manual > date_sub - Subtracts an amount of days, months, years, hours, minutes and seconds from a DateTime object

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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.

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