DateTime::sub

date_sub

(PHP 5 >= 5.3.0, PHP 7)

DateTime::sub -- date_sub — 对一个 DateTime 对象减去一定量的日、月、年、小时、分钟和秒。

说明

面向对象风格

public DateTime::sub ( DateInterval $interval ) : DateTime

过程化风格

date_sub ( DateTime $object , DateInterval $interval ) : DateTime

DateInterval 对象，表示要减去多少量的时间。

参数

object: 仅过程化风格：由 date_create() 返回的 DateTime 类型的对象。此函数会修改这个对象。
interval: DateInterval 对象

返回值

返回被修改的 DateTime 对象，或者在失败时返回 FALSE.

范例

Example #1 DateTime::sub() 例程

面向对象风格


<?php
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('P10D'));
echo $date->format('Y-m-d') . "\n";
?>

过程化风格


<?php
$date = date_create('2000-01-20');
date_sub($date, date_interval_create_from_date_string('10 days'));
echo date_format($date, 'Y-m-d');
?>

以上例程会输出：

2000-01-10

Example #2 Further DateTime::sub() 例程


<?php
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('PT10H30S'));
echo $date->format('Y-m-d H:i:s') . "\n";

$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('P7Y5M4DT4H3M2S'));
echo $date->format('Y-m-d H:i:s') . "\n";
?>

以上例程会输出：

2000-01-19 13:59:30
1992-08-15 19:56:58

Example #3 减去一定量的月份的时候需要注意


<?php
$date = new DateTime('2001-04-30');
$interval = new DateInterval('P1M');

$date->sub($interval);
echo $date->format('Y-m-d') . "\n";

$date->sub($interval);
echo $date->format('Y-m-d') . "\n";
?>

以上例程会输出：

2001-03-30
2001-03-02

注释

在 PHP 5.2 之后，可以使用 DateTime::modify() 作为替代。

参见

DateTime::add() - 给一个 DateTime 对象增加一定量的天，月，年，小时，分钟以及秒。
DateTime::diff() - Returns the difference between two DateTime objects
DateTime::modify() - 修改日期时间对象的值

User Contributed Notes

david_newey at cable dot comcast dot com 26-Dec-2018 06:06


As noted above when subtracting months, results can be suspect. I needed to create an array of end of month dates for 6 months starting at Oct and going back.  Using:

<?php



//Instantiate array

$dateset = [];



//Create new date object

$date = new \DateTime('2018-10-31);

//Add to array

$dateset[] = $date->format('Y-m-d');



//Go back 5 months

$nbr = 6;

for($i = 1; $i < $nbr; $i++){

    $date->sub(new \DateInterval('P1M'));

    $dateset[] = $date->format('Y-m-d');

}

?>



Results in:



array:6 [▼

  0 => "2018-10-31"

  1 => "2018-10-01"

  2 => "2018-09-01"

  3 => "2018-08-01"

  4 => "2018-07-01"

  5 => "2018-06-01"

]



However, using ->modify("last day of last month") accurately gives month ending dates:



<?php



//Instantiate array

$dateset = [];



//Create new date object

$date = new \DateTime('2018-10-31);

//Add to array

$dateset[] = $date->format('Y-m-d');



//Go back 5 months

$nbr = 6;

for($i = 1; $i < $nbr; $i++){

    $date->modify('last day of last month');

    $dateset[] = $date->format('Y-m-d');

}

?>



Results in:



array:6 [▼

  0 => "2018-10-31"

  1 => "2018-09-30"

  2 => "2018-08-31"

  3 => "2018-07-31"

  4 => "2018-06-30"

  5 => "2018-05-31"

]

Anonymous 08-Oct-2017 12:00


yeoo



is this in utc or not i have no clue

soerenklein98 at gmail dot com 16-Sep-2015 06:17


Keep in mind that this method can work incorrectly when the daylight saving time is changing to wintertime.

An example is when you have got a loop and your subtracting 1 hour from your time in it. On one day you are actually subtracting 1 hour and the computer adds 1 - the infinitive loop is created.

And on the other day you would actually subtract 2 hours.

itonohito 05-Apr-2014 07:25


When trying to pass daylight saving state change time, sub() works incorrectly.



$t = new DateTime( '2014-03-30 02:00:00' );

$t->add( new DateInterval('PT1H') );

echo $->format('Y-m-d H:i:s');



output will be: '2014-03-30 04:00:00'.



 Well, it's ok because at 3:00 a.m. daylight saving time begins in my country, so after  02:59:59 must be 04:00:00.



But if I try to subtract time:



$t = new DateTime( '2014-03-30 04:00:00' );

$t->sub( new DateInterval('PT1H') );

echo $->format('Y-m-d H:i:s');



output will be: '2014-03-30 04:00:00'.



Yes, completely the same, not '2014-03-30 02:00:00' as it should be.

Anonymous 01-Feb-2011 02:15


Note that the sub() and add() methods will modify the value of the object you're calling the method on! This is very untypical for a method that returns a value of its own type. You could misunderstand it that the method would return a new instance with the modified value, but in fact it modifies itself! This is undocumented here. (Only a side note on procedural style mentions it, but it obviously does not apply to object oriented style.)

php at keith tyler dot com 30-Nov-2010 06:53


If you use diff() after sub(), the effects of the sub() will be repeated on the date object.



It doesn't matter if the object is the one diffed or doing the diffing (i.e. which object you call diff() from).



<?php

$today = new DateTime();

$newdate = new DateTime();



print_r($newdate);

$newdate->sub(new DateInterval("PT1S"));

print_r($newdate);

$s = $newdate->diff($today);

print_r($newdate);

$s = $today->diff($newdate);

print_r($newdate);

$s = $today->diff($newdate);

print_r($newdate);

?>



Prints:



DateTime Object

(

    [date] => 2010-11-30 18:43:48

    [timezone_type] => 3

    [timezone] => America/Los_Angeles

)

DateTime Object

(

    [date] => 2010-11-30 18:43:47

    [timezone_type] => 3

    [timezone] => America/Los_Angeles

)

DateTime Object

(

    [date] => 2010-11-30 18:43:46

    [timezone_type] => 3

    [timezone] => America/Los_Angeles

)

DateTime Object

(

    [date] => 2010-11-30 18:43:45

    [timezone_type] => 3

    [timezone] => America/Los_Angeles

)

DateTime Object

(

    [date] => 2010-11-30 18:43:44

    [timezone_type] => 3

    [timezone] => America/Los_Angeles

)



Note that using add() instead of sub() does NOT have the same effect.



This is particularly undesirable -- in this example you make a datetime, use sub() to make it a relative time in the past, and then date->diff() to confirm the difference. But the diff() inadvertendly makes the difference 2x.