date_sub

(PHP 5 >= 5.3.0, PHP 7)

说明

此函数是该函数的别名： DateTime::sub()

User Contributed Notes

fpradove at gmail dot com 07-Apr-2018 12:05


In version 5.6.31 the variable $ today is passed by reference in the function date_sub () and the interval is also applied



<?php 

$today = date_create(date('Y-m-d'));

$yesterday = date_sub($today, date_interval_create_from_date_string("1 days"));



        echo var_dump($today);

        echo var_dump($yesterday)

?>

vitkor dot kruglikov at gmail dot com 15-Jun-2015 08:55


It can be unclear for someone how to use this function.

Here is the example:



$date=date_create("2013-03-15");

date_sub($date,date_interval_create_from_date_string("40 days"));

echo date_format($date,"Y-m-d");

Thomas LEDUC 04-Feb-2015 11:14


To glue to the OOP, it's better to use it with DateInterval::createFromDateString



<?php 

$dateB = new DateTime('2020-12-20'); 

$dateA = $dateB->sub(DateInterval::createFromDateString('10 days'));

?>



more detail here :

<?php

public static function createFromDateString ($time) {}

?>

Jonathan Poissant 21-Jan-2011 07:26


You cannot replace date_sub('2000-01-20') by DateTime::sub('2000-01-20') because DateTime::sub is not static. You have to create the DateTime object first.





Example:





<?php $dateA = date_sub('2000-01-20', date_interval_create_from_date_string('10 days')); ?>





will be replace by


<?php


$dateB = new DateTime('2000-01-20');


$dateA = $dateB->sub(date_interval_create_from_date_string('10 days'));


?>