Adding days to $Date in PHP

Question

I have a date returned as part of a MySQL query in the form 2010-09-17.

I would like to set the variables $Date2 to $Date5 as follows:

$Date2 = $Date + 1

$Date3 = $Date + 2

etc., so that it returns 2010-09-18, 2010-09-19, etc.

I have tried

date('Y-m-d', strtotime($Date. ' + 1 day'))

but this gives me the date before $Date.

What is the correct way to get my Dates in the format form 'Y-m-d' so that they may be used in another query?

possible duplicate of [PHP Calculating future date by adding days to a variable date](http://stackoverflow.com/questions/2692679/php-calculating-future-date-by-adding-days-to-a-variable-date) — Gordon, Sep 16 '10 at 14:36

score 563 · Accepted Answer · answered Sep 16 '10 at 14:46

563

All you have to do is use days instead of day like this:

<?php
$Date = "2010-09-17";
echo date('Y-m-d', strtotime($Date. ' + 1 days'));
echo date('Y-m-d', strtotime($Date. ' + 2 days'));
?>

And it outputs correctly:

2010-09-18
2010-09-19

answered Sep 16 '10 at 14:46

shamittomar

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8

It outputs correctly using just `day` as well... `date("Y-m-d", strtotime('2010-09-17 + 1 day'))` -> `2010-09-18`, `date("Y-m-d", strtotime('2010-09-17 + 2 day'))` -> `2010-09-19` – Daniel Vandersluis Sep 16 '10 at 14:50
Just tried it this way and it worked. I still have no idea why I got a value one day before my start date when I initially tried it. – Istari Sep 16 '10 at 15:30
What if I had a variable for de number of days to add? – CMartins Jun 14 '14 at 10:53
4

@Carlos echo date("Y-m-d", strtotime($Date.' + '.$variable.' days')); – Pradeep Kumar Feb 04 '16 at 10:31
Nope this method doesn't works at all. It is returning some date of 2004. – gegobyte Feb 05 '16 at 10:16

Daniel Vandersluis · Answer 2 · 2010-09-16T14:57:10.633

110

If you're using PHP 5.3, you can use a DateTime object and its add method:

$Date1 = '2010-09-17';
$date = new DateTime($Date1);
$date->add(new DateInterval('P1D')); // P1D means a period of 1 day
$Date2 = $date->format('Y-m-d');

Take a look at the DateInterval constructor manual page to see how to construct other periods to add to your date (2 days would be 'P2D', 3 would be 'P3D', and so on).

Without PHP 5.3, you should be able to use strtotime the way you did it (I've tested it and it works in both 5.1.6 and 5.2.10):

$Date1 = '2010-09-17';
$Date2 = date('Y-m-d', strtotime($Date1 . " + 1 day"));
// var_dump($Date2) returns "2010-09-18"

edited Sep 16 '10 at 14:57

answered Sep 16 '10 at 14:26

Daniel Vandersluis

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Sorry Daniel, but the webserver runs php 5.2.6, so this won't work – Istari Sep 16 '10 at 14:27
2

`DateTime::construct` uses the same mechanism as `strtotime` to parse the date, so you can also do `new DateTime("+1 day $date")` which would not require 5.3 – Gordon Sep 16 '10 at 14:37
This is the actual answer for php 5.5+ – TheLegendaryCopyCoder Dec 08 '18 at 16:41
1

@Steven ► https://meta.stackoverflow.com/questions/294938/how-can-i-undo-my-vote – Jose Manuel Abarca Rodríguez Sep 01 '22 at 20:54
@JoseManuelAbarcaRodríguez - Heh yea I tried that but apparently my vote is locked unless this answer is edited (according to SO) – Steven Sep 01 '22 at 21:36

Omn · Answer 3 · 2015-02-18T01:06:49.273

40

From PHP 5.2 on you can use modify with a DateTime object:

http://php.net/manual/en/datetime.modify.php

$Date1 = '2010-09-17';
$date = new DateTime($Date1);
$date->modify('+1 day');
$Date2 = $date->format('Y-m-d');

Be careful when adding months... (and to a lesser extent, years)

edited Feb 18 '15 at 01:06

answered Jan 28 '15 at 18:25

Omn

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accepted answer just gives answer for 15000+ days where this answer gives answer when you add 25000+ days too.. – Pradeep Kumar Feb 04 '16 at 11:07
You can also query this on a DateTimeImmutable() and you get a modified copy without risk of altering the original. – Xavi Montero May 07 '18 at 00:53

Madhu V Rao · Answer 4 · 2015-09-24T04:24:28.910

23

Here is a small snippet to demonstrate the date modifications:

$date = date("Y-m-d");
//increment 2 days
$mod_date = strtotime($date."+ 2 days");
echo date("Y-m-d",$mod_date) . "\n";

//decrement 2 days
$mod_date = strtotime($date."- 2 days");
echo date("Y-m-d",$mod_date) . "\n";

//increment 1 month
$mod_date = strtotime($date."+ 1 months");
echo date("Y-m-d",$mod_date) . "\n";

//increment 1 year
$mod_date = strtotime($date."+ 1 years");
echo date("Y-m-d",$mod_date) . "\n";

edited Sep 24 '15 at 04:24

answered Feb 25 '13 at 06:07

Madhu V Rao

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can you please see my related (quesiton)[https://stackoverflow.com/q/61421783/8919244]? – Moeez Apr 25 '20 at 07:23

score 11 · Answer 5 · answered Nov 22 '15 at 18:46

11

Here has an easy way to solve this.

<?php
   $date = "2015-11-17";
   echo date('Y-m-d', strtotime($date. ' + 5 days'));
?>

Output will be:

2015-11-22

Solution has found from here - How to Add Days to Date in PHP

answered Nov 22 '15 at 18:46

JoyGuru

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score 9 · Answer 6 · edited Apr 06 '15 at 14:41

You can also use the following format

strtotime("-3 days", time());
strtotime("+1 day", strtotime($date));

You can stack changes this way:

strtotime("+1 day", strtotime("+1 year", strtotime($date)));

Note the difference between this approach and the one in other answers: instead of concatenating the values +1 day and <timestamp>, you can just pass in the timestamp as the second parameter of strtotime.

score 9 · Answer 7 · edited Jan 27 '23 at 02:24

9

Using a variable for Number of days

$myDate = "2014-01-16";
$nDays = 16;
$newDate = strtotime($myDate . '+ '.$nDays.' days');
echo new Date('d/m/Y', $newDate); //format new date

edited Jan 27 '23 at 02:24

Ahmad Gulzar

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answered Jun 18 '14 at 08:25

CMartins

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What is "`$soma`"? Some weird typo of "`$newDate`"? – Peter Mortensen Jan 26 '23 at 14:20

score 6 · Answer 8 · edited Mar 29 '19 at 13:08

6

Here is the simplest solution to your query

$date=date_create("2013-03-15"); // or your date string
date_add($date,date_interval_create_from_date_string("40 days"));// add number of days 
echo date_format($date,"Y-m-d"); //set date format of the result

edited Mar 29 '19 at 13:08

Fury

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answered Sep 21 '16 at 18:41

Engr Atiq

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score 2 · Answer 9 · answered Oct 08 '21 at 07:42

This works. You can use it for days, months, seconds and reformat the date as you require

public function reformatDate($date, $difference_str, $return_format)
{
    return date($return_format, strtotime($date. ' ' . $difference_str));
}

Examples

echo $this->reformatDate('2021-10-8', '+ 15 minutes', 'Y-m-d H:i:s');
echo $this->reformatDate('2021-10-8', '+ 1 hour', 'Y-m-d H:i:s');
echo $this->reformatDate('2021-10-8', '+ 1 day', 'Y-m-d H:i:s');

score 1 · Answer 10 · answered Apr 28 '21 at 19:49

To add a certain number of days to a date, use the following function.

function add_days_to_date($date1,$number_of_days){
    /*
    //$date1 is a string representing a date such as '2021-04-17 14:34:05'
    //$date1 =date('Y-m-d H:i:s'); 
    // function date without a secrod argument returns the current datetime as a string in the specified format
    */
    $str =' + '. $number_of_days. ' days';
    $date2= date('Y-m-d H:i:s', strtotime($date1. $str));
    return $date2; //$date2 is a string
}//[end function]

score 0 · Answer 11 · edited Dec 14 '12 at 11:08

0

All have to use bellow code:

$nday = time() + ( 24 * 60 * 60);    
echo 'Now:       '. date('Y-m-d') ."\n";    
echo 'Next Day: '. date('Y-m-d', $nday) ."\n";

edited Dec 14 '12 at 11:08

Sumit Singh

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answered Dec 14 '12 at 10:51

vikasgore

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You shouldn't do it this way, eventually you will have problems with leap seconds – Nathan Aug 01 '13 at 06:07

Pikamander2 · Answer 12 · 2022-05-24T12:35:52.793

Another option is to convert your date string into a timestamp and then add the appropriate number of seconds to it.

$datetime_string = '2022-05-12 12:56:45';
$days_to_add = 1;
$new_timestamp = strtotime($datetime_string) + ($days_to_add * 60 * 60 * 24);

After which, you can use one of PHP's various date functions to turn the timestamp into a date object or format it into a human-readable string.

$new_datetime_string = date('Y-m-d H:i:s', $new_timestamp);

Adding days to $Date in PHP

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