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phpdate
Istari
Istari

Reputation: 3983

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?

Upvotes: 291

Views: 656573

Answers (13)

lovelesh singh
lovelesh singh

Reputation: 23

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

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

Upvotes: -2

CMartins
CMartins

Reputation: 3293

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

Upvotes: 9

Pikamander2
Pikamander2

Reputation: 8319

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);

Upvotes: 1

Dankyi Anno Kwaku
Dankyi Anno Kwaku

Reputation: 1293

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');

Upvotes: 2

Meisam Rasouli
Meisam Rasouli

Reputation: 321

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]

Upvotes: 1

Engr Atiq
Engr Atiq

Reputation: 163

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

Upvotes: 6

JoyGuru
JoyGuru

Reputation: 1833

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

Upvotes: 12

Madhu V Rao
Madhu V Rao

Reputation: 937

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";

Upvotes: 23

SztupY
SztupY

Reputation: 10546

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.

Upvotes: 9

Omn
Omn

Reputation: 3080

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)

Upvotes: 41

vikasgore
vikasgore

Reputation: 17

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";

Upvotes: 0

Daniel Vandersluis
Daniel Vandersluis

Reputation: 94284

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"

Upvotes: 113

shamittomar
shamittomar

Reputation: 46702

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

Upvotes: 577

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