[php] Adding days to $Date in PHP

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?

This question is related to php date

The answer is


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";

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"

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";

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

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


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.


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)


Using a variable for Number of days

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