How to add 1 month on a date without skipping i.e. february [duplicate]

Answer 1

OK - as far as I can tell, you still haven't clearly defined what you mean by "month".

But here's one more function that might help:

http://php.net/manual/en/function.getdate.php

/* Function to find the first and last day of the month from the given date.
*
* Author Binu v Pillai            binupillai2003@yahoo.com
* @Param            String             yyyy-mm-dd
*
*/
function findFirstAndLastDay($anyDate)
{
    //$anyDate            =    '2009-08-25';    // date format should be yyyy-mm-dd
    list($yr,$mn,$dt)    =    split('-',$anyDate);    // separate year, month and date
    $timeStamp            =    mktime(0,0,0,$mn,1,$yr);    //Create time stamp of the first day from the give date.
    $firstDay            =     date('D',$timeStamp);    //get first day of the given month
    list($y,$m,$t)        =     split('-',date('Y-m-t',$timeStamp)); //Find the last date of the month and separating it
    $lastDayTimeStamp    =    mktime(0,0,0,$m,$t,$y);//create time stamp of the last date of the give month
    $lastDay            =    date('D',$lastDayTimeStamp);// Find last day of the month
    $arrDay                =    array("$firstDay","$lastDay"); // return the result in an array format.

    return $arrDay;
}

Answer 2

You have several alternatives besides DateInterval.

Here are examples that use strtotime():

http://www.brightcherry.co.uk/scribbles/php-adding-and-subtracting-dates/

// Subtracting days from a date
$date = "1998-08-14";
$newdate = strtotime ( '-3 day' , strtotime ( $date ) ) ;
$newdate = date ( 'Y-m-j' , $newdate );

echo $newdate;

...

// Subtracting months from a date
$date = "1998-08-14";
$newdate = strtotime ( '-3 month' , strtotime ( $date ) ) ;
$newdate = date ( 'Y-m-j' , $newdate );

echo $newdate;

Here's are some links for DateInterval:

• http://php.net/manual/en/dateinterval.construct.php

• What can go wrong when adding months with a DateInterval and DateTime::add?

Q: Exactly how do you define a "month"?

Def#1): a "month" is a "month" - regardless of #/days

Def#2): a "month" is 30 days (for example)

Def#3): a "month" is the #/days between the 1st Monday of subsequent months

etc. etc

Q: What exactly is your "definition"?

Answer 3

It seems that there is no ready function for it, so I wrote it by myself. This should solve my problem. Thanks anyway for your answers and comments. If you will find some errors, please provide in the comments.

This function calculates the month and the year I will end in after adding some month. Then it checks if the date is right. If not, we have the case that the day is not in the target month, so we take the last day in the month instead. Tested with PHP 5.3.10.

<?php
$monthToAdd = 1;

$d1 = DateTime::createFromFormat('Y-m-d H:i:s', '2011-01-30 15:57:57');

$year = $d1->format('Y');
$month = $d1->format('n');
$day = $d1->format('d');

$year += floor($monthToAdd/12);
$monthToAdd = $monthToAdd%12;
$month += $monthToAdd;
if($month > 12) {
    $year ++;
    $month = $month % 12;
    if($month === 0)
        $month = 12;
}

if(!checkdate($month, $day, $year)) {
    $d2 = DateTime::createFromFormat('Y-n-j', $year.'-'.$month.'-1');
    $d2->modify('last day of');
}else {
    $d2 = DateTime::createFromFormat('Y-n-d', $year.'-'.$month.'-'.$day);
}
$d2->setTime($d1->format('H'), $d1->format('i'), $d1->format('s'));
echo $d2->format('Y-m-d H:i:s');