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?

Oberstone

possible duplicate of PHP Calculating future date by adding days to a variable date – Metric

Answer 1

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

Beauharnais

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 – Pacifically

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. – Oberstone

What if I had a variable for de number of days to add? – Ranique

@Carlos echo date("Y-m-d", strtotime($Date.' + '.$variable.' days')); – Frohman

Nope this method doesn't works at all. It is returning some date of 2004. – Keeler

Answer 2

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"

Pacifically

Sorry Daniel, but the webserver runs php 5.2.6, so this won't work – Oberstone

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 – Metric

This is the actual answer for php 5.5+ – Trough

@Steven ► meta.https://mcmap.net/q/109958/-web-page-designers – Masaccio

@JoseManuelAbarcaRodríguez - Heh yea I tried that but apparently my vote is locked unless this answer is edited (according to SO) – Conciliate

Answer 3

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)

Hysterogenic

accepted answer just gives answer for 15000+ days where this answer gives answer when you add 25000+ days too.. – Frohman

You can also query this on a DateTimeImmutable() and you get a modified copy without risk of altering the original. – Fissure

Answer 4

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";

Potation

Answer 5

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

Conveyancer

Answer 6

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.

Lobworm

Answer 7

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 

Ranique

What is "$soma"? Some weird typo of "$newDate"? – Halo

Answer 8

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

Fingering

Answer 9

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');

Arceliaarceneaux

Answer 10

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]

Fiddlehead

Answer 11

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";

Unshaped

You shouldn't do it this way, eventually you will have problems with leap seconds – Ormazd

Answer 12

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);

Hageman

Answer 13

We can implement simply add or subtract days in the current date in php

date('Y-m-d' , strtotime('-90days'));

Bush

OP says: "I have a date returned as part of a MySQL query in the form 2010-09-17." and they want to add days to that date. Please only answer the asked question. – Tove