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");
Описание
Функция — псевдоним функции: DateTime::sub()
Смотрите также
- DateTimeImmutable::sub() - Вычитает переданное количество дней, месяцев, лет, часов, минут и секунд
- DateTime::sub() - Вычитает дни, месяцы, годы, часы, минуты и секунды из объекта DateTime
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User Contributed Notes 4 notes
vitkor dot kruglikov at gmail dot com ¶
9 years ago
fpradove at gmail dot com ¶
6 years ago
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)
?>
Thomas LEDUC ¶
9 years ago
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 ¶
13 years ago
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'));
?>
