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.)
DateTime::sub
date_sub
(PHP 5 >= 5.3.0, PHP 7)
DateTime::sub -- date_sub — Subtracts an amount of days, months, years, hours, minutes and seconds from a DateTime object
Description
Object oriented style
Procedural style
Subtracts the specified DateInterval object from the specified DateTime object.
Parameters
-
object -
Procedural style only: A DateTime object returned by date_create(). The function modifies this object.
-
interval -
A DateInterval object
Return Values
Returns the DateTime object for method chaining or FALSE on failure.
Examples
Example #1 DateTime::sub() example
Object oriented style
<?php
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('P10D'));
echo $date->format('Y-m-d') . "\n";
?>
Procedural style
<?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');
?>
The above examples will output:
2000-01-10
Example #2 Further DateTime::sub() examples
<?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";
?>
The above example will output:
2000-01-19 13:59:30 1992-08-15 19:56:58
Example #3 Beware when subtracting months
<?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";
?>
The above example will output:
2001-03-30 2001-03-02
Notes
DateTime::modify() is an alternative when using PHP 5.2.
See Also
- DateTime::add() - Adds an amount of days, months, years, hours, minutes and seconds to a DateTime object
- DateTime::diff() - Returns the difference between two DateTime objects
- DateTime::modify() - Alters the timestamp
add a note
User Contributed Notes 6 notes
Anonymous ¶
8 years ago
php at keith tyler dot com ¶
8 years ago
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.
david_newey at cable dot comcast dot com ¶
9 months ago
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 [â–¼
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 [â–¼
0 => "2018-10-31"
1 => "2018-09-30"
2 => "2018-08-31"
3 => "2018-07-31"
4 => "2018-06-30"
5 => "2018-05-31"
]
itonohito ¶
5 years ago
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.
soerenklein98 at gmail dot com ¶
4 years ago
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.
