PHP: generating an end date with user specified start date and duration - php

Question

I am trying to create a function that determines the end date of an advert based on the start date and duration parameter passed by a user.
Example:
If user specify start date as 5th June 2013 and duration as 45 days:
$ad_startdate = '2013-06-05';
$ad_duration = 45;
The function should automatically get the end date which should be 20th July 2013:
$ad_end_date = '2013-07-20';
Pls note that to make it easy to generate the end date, I've assigned the variable for months a constant value which will be 30 days. Whether it's february or november or a leap year, every month has a fixed variable value of 30.
I was trying to come up with something around this but just cant figure it out.
$ad_startdate = '2013-06-05';
$ad_duration = 45;
// End date should be 2013-07-20
function getAdvertEndDate ($ad_startdate, $ad_duration){
//Add up start date with duration
$end_date = strtotime($ad_startdate) + $ad_duration;
return $end_date;
}
I have browsed through SO questions just to see if anyone has something around this but the answered ones are so different from mine challenge.
Would be very grateful getting help with this.

Answer 1

function getAdvertEndDate ($ad_startdate, $ad_duration){
return date("Y-m-d", strtotime($ad_startdate) + ($ad_duration * 86400));
}
Use like so:
$endDate = getAdvertEndDate("2013-04-08", 40);

Answer 2

PHP >= 5.3.0 Object oriented style
$date = DateTime::createFromFormat('Y-m-d', '2013-06-05');
$date->add(new DateInterval('P45D'));
echo $date->format('Y-m-d') . "\n";
Or Procedural style
$date = date_create('2013-06-05');
date_add($date, date_interval_create_from_date_string('45 days'));
echo date_format($date, 'Y-m-d');
Result:
2013-07-20
Code:
function getAdvertEndDate ($ad_startdate, $ad_duration){
$date = DateTime::createFromFormat('Y-m-d', $ad_startdate);
$date->add(new DateInterval('P'.$ad_duration.'D'));
return $date->format('Y-m-d');
}
For PHP < 5.3 use strtotime():
function getAdvertEndDate ($ad_startdate, $ad_duration){
//Add up start date with duration
return date('Y-m-d', strtotime($ad_startdate. " + $ad_duration days"));
}
echo getAdvertEndDate('2013-06-05', '45'); // 2013-07-20
http://www.php.net/manual/en/datetime.add.php

Answer 3

Try this code
$date = '2013-06-05';
$date1 = strtotime($date);
$date2 = strtotime('+45 day',$date1);
echo date('Y-m-d', $date2);

Answer 4

The native strtotime() function does this work.

Answer 5

Use this:
$ad_startdate = '2013-06-05';
$ad_duration = 45;
$dateArray = explode('-', $ad_startdate);
$newDate = date('Y-m-d', strtotime('+ ' . $ad_duration . ' days', mktime(0, 0, 0, $dateArray[1], $dateArray[2], $dateArray[0]));
If you're using strtotime, you cant use the date format you've specified, as if using - seperators, strtotime() expects the format differently.
From PHP.net
Note:
Dates in the m/d/y or d-m-y formats are disambiguated by looking at the separator between the various components: if the separator is a slash (/), then the American m/d/y is assumed; whereas if the separator is a dash (-) or a dot (.), then the European d-m-y format is assumed.
To avoid potential ambiguity, it's best to use ISO 8601 (YYYY-MM-DD) dates or DateTime::createFromFormat() when possible.