How can I calculate End date automatically in Php ?
When I enter current date it will be auto calculate 16moths or 480days.
duration between current and end date 16month and days.
<?php
$Date = "2021-09-27";
echo date('Y-m-d', strtotime($Date. ' + 480 days'));
?>
try this solution:
$currentDate = date('Y-m-d');
$endDate = date('Y-m-d', strtotime($currentDate. ' + 480 days'));
echo $currentDate;
echo '<br>';
echo $endDate;
here is the solution
$time = new DateTime(date('Y:m:d H:i:s'));
$time->add(new DateInterval('P10D'));
echo $endtime = $time->format('Y:m:d H:i:s');
for more info visit: https://www.php.net/manual/en/datetime.add.php
Related
I have this code
<?php
$date =date(Y-m-d);
$day = 5;
$newdate= $date+$day
echo "today is:"$date;
echo "<br> and after 5 days is :"$newdate;
?>
I want the result is
today is :2016-11-2
and after 5 days is : 2016-11-7
Try this
$date = new DateTime(); // Creates new DatimeTime for today
$newdate = $date->modify( '+ 5 days' ); // Adds 5 days
echo $newdate->format( 'Y-m-d' ); // Echo and format the newdate to the wanted format
It should help you:
echo date('Y-m-d', strtotime($date. ' + 5 days'));
So it will be like follows:
<?php
$date = date('Y-m-d');
$newdate = date('Y-m-d', strtotime($date.' + 5 days'));
echo "today is: $date";
echo "<br> and after 5 days is: $newdate";
?>
You can use strtotime() function to add days to current date. Please see the below :
<?php
$date =date("Y-m-d");
$day = 5;
$newdate=date('Y-m-d', strtotime("+$day days"));
echo "today is:".$date;
echo "<br> and after 5 days is :".$newdate;
?>
it may help you
$date = "Mar 03, 2016";
$date = strtotime($date);
$date = strtotime("+7 day", $date);
echo date('M d, Y', $date);
I read about this but does not working for me. Here is my code:
$today = date_create()->format("d/m/Y"); // Today is 25/04/2013
$num_days = GetNumberOfdays();
$end_date = date("d/m/Y", strtotime($today . " + $num_days days"));
The value that I get from $end_date is 31/12/1969. What am I doing wrong?
Try this instead:
$end_date = date("d/m/Y", strtotime("+ $num_days days", time()));
EDIT: I changed the $today variable to just time() which is essentially getting you the same information if you're just looking for today's date.
From what it looks like you're trying to do, you don't even need $today (as it defaults to now if date is not supplied), so you could just do eg:
$end_date = date("d/m/Y", strtotime("+ 5 days"));
echo $end_date;
result would be
30/04/2013
if you want to provide a date, you need the parameters the other way round, as per the manual:
strtotime ( string $time [, int $now = time() ] )
date_create() return a DateTime object.
You could use DateTime::modify method.
$date = new \DateTime(); // Defaults to Today
$num_days = 123;
$date->add(
new \DateInterval('P' . $num_days . 'D')
);
echo $date->format('d-M-Y');
$today = date_create()->format("d/m/Y"); // Today is 25/04/2013
$num_days = date_create()->format("d");
echo $end_date = date("d/m/Y", strtotime(" + $num_days days"));
<?
// note change of $today format
$today = date_create()->format("d-m-Y"); // Today is 25-04-2013
$num_days = GetNumberOfdays();
$end_date = date("d/m/Y", strtotime("+" . $num_days . " days", strtotime($today)));
?>
I have a PHP date in the form of 2013-01-22 and I want to get tomorrows date in the same format, so for example 2013-01-23.
How is this possible with PHP?
Use DateTime
$datetime = new DateTime('tomorrow');
echo $datetime->format('Y-m-d H:i:s');
Or:
$datetime = new DateTime('2013-01-22');
$datetime->modify('+1 day');
echo $datetime->format('Y-m-d H:i:s');
Or:
$datetime = new DateTime('2013-01-22');
$datetime->add(new DateInterval("P1D"));
echo $datetime->format('Y-m-d H:i:s');
Or in PHP 5.4+:
echo (new DateTime('2013-01-22'))->add(new DateInterval("P1D"))
->format('Y-m-d H:i:s');
$tomorrow = date("Y-m-d", strtotime('tomorrow'));
or
$tomorrow = date("Y-m-d", strtotime("+1 day"));
Help Link: STRTOTIME()
Since you tagged this with strtotime, you can use it with the +1 day modifier like so:
$tomorrow_timestamp = strtotime('+1 day', strtotime('2013-01-22'));
That said, it's a much better solution to use DateTime.
<? php
//1 Day = 24*60*60 = 86400
echo date("d-m-Y", time()+86400);
?>
echo date ('Y-m-d',strtotime('+1 day', strtotime($your_date)));
Use DateTime:
To get tomorrow from now :
$d = new DateTime('+1day');
$tomorrow = $d->format('d/m/Y h.i.s');
echo $tomorrow;
Results : 28/06/2017 08.13.20
To get tomorrow from a date :
$d = new DateTime('2017/06/10 08.16.35 +1day')
$tomorrow = $d->format('d/m/Y h.i.s');
echo $tomorrow;
Results : 11/06/2017 08.16.35
Hope it helps!
/**
* get tomorrow's date in the format requested, default to Y-m-d for MySQL (e.g. 2013-01-04)
*
* #param string
*
* #return string
*/
public static function getTomorrowsDate($format = 'Y-m-d')
{
$date = new DateTime();
$date->add(DateInterval::createFromDateString('tomorrow'));
return $date->format($format);
}
By strange it can seem it works perfectly fine: date_create( '2016-02-01 + 1 day' );
echo date_create( $your_date . ' + 1 day' )->format( 'Y-m-d' );
Should do it
here's working function
function plus_one_day($date){
$date2 = formatDate4db($date);
$date1 = str_replace('-', '/', $date2);
$tomorrow = date('Y-m-d',strtotime($date1 . "+1 days"));
return $tomorrow; }
$date = '2013-01-22';
$time = strtotime($date) + 86400;
echo date('Y-m-d', $time);
Where 86400 is the # of seconds in a day.
I want to add 5 minutes to this date: 2011-04-8 08:29:49
$date = '2011-04-8 08:29:49';
When I use strtotime I am always getting 1970-01-01 08:33:31
How do I add correctly 5 minutes to 2011-04-8 08:29:49?
$date = '2011-04-8 08:29:49';
$currentDate = strtotime($date);
$futureDate = $currentDate+(60*5);
$formatDate = date("Y-m-d H:i:s", $futureDate);
Now, the result is 2011-04-08 08:34:49 and is stored inside $formatDate
Enjoy! :)
Try this:
echo date('Y-m-d H:i:s', strtotime('+5 minutes', strtotime('2011-04-8 08:29:49')));
$expire_stamp = date('Y-m-d H:i:s', strtotime("+5 min"));
$now_stamp = date("Y-m-d H:i:s");
echo "Right now: " . $now_stamp;
echo "5 minutes from right now: " . $expire_stamp;
Results in:
2012-09-30 09:00:03
2012-09-30 09:05:03
$date = '2011-04-8 08:29:49';
$newDate = date("Y-m-d H:i:s",strtotime($date." +5 minutes"))
For adding
$date = new DateTime('2014-02-20 14:20:00');
$date->add(new DateInterval('P0DT0H5M0S'));
echo $date->format('Y-m-d H:i:s');
It add 5minutes
For subtracting
$date = new DateTime('2014-02-20 14:20:00');
$date->sub(new DateInterval('P0DT0H5M0S'));
echo $date->format('Y-m-d H:i:s');
It subtract 5 minutes
If i'm right in thinking.
If you convert your date to a unix timestamp via strtotime(), then just add 300 (5min * 60 seconds) to that number.
$timestamp = strtotime($date) + (5*60)
Hope this helps
more illustrative for simple and clear solution
$date = '2011-04-8 08:29:49';
$newtimestamp = strtotime($date. ' + 5 minute');//gets timestamp
//convert into whichever format you need
$newdate = date('Y-m-d H:i:s', $newtimestamp);//it prints 2011-04-08 08:34:49
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?
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
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"
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)
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",$mod_date) . "\n";
You can also use the following format
strtotime("-3 days", time());
strtotime("+1 day", strtotime($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.
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
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
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($date,"Y-m-d"); //set date format of the result
This works. You can use it for days, months, seconds and reformat the date as you require
public function reformatDate($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 $this->reformatDate('2021-10-8', '+ 1 day', 'Y-m-d H:i:s');
To add a certain number of days to a date, use the following function.
function add_days_to_date($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]
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";
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);