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?
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
Source: Stackoverflow.com