increment date by one month - php

Question

Let's say I have a date in the following format: 2010-12-11 (year-mon-day)
With PHP, I want to increment the date by one month, and I want the year to be automatically incremented, if necessary (i.e. incrementing from December 2012 to January 2013).
Regards.

Answer 1

$time = strtotime("2010.12.11");
$final = date("Y-m-d", strtotime("+1 month", $time));
// Finally you will have the date you're looking for.

Answer 2

I needed similar functionality, except for a monthly cycle (plus months, minus 1 day). After searching S.O. for a while, I was able to craft this plug-n-play solution:
function add_months($months, DateTime $dateObject)
{
$next = new DateTime($dateObject->format('Y-m-d'));
$next->modify('last day of +'.$months.' month');
if($dateObject->format('d') > $next->format('d')) {
return $dateObject->diff($next);
} else {
return new DateInterval('P'.$months.'M');
}
}
function endCycle($d1, $months)
{
$date = new DateTime($d1);
// call second function to add the months
$newDate = $date->add(add_months($months, $date));
// goes back 1 day from date, remove if you want same day of month
$newDate->sub(new DateInterval('P1D'));
//formats final date to Y-m-d form
$dateReturned = $newDate->format('Y-m-d');
return $dateReturned;
}
Example:
$startDate = '2014-06-03'; // select date in Y-m-d format
$nMonths = 1; // choose how many months you want to move ahead
$final = endCycle($startDate, $nMonths); // output: 2014-07-02

Answer 3

Use DateTime::add.
$start = new DateTime("2010-12-11", new DateTimeZone("UTC"));
$month_later = clone $start;
$month_later->add(new DateInterval("P1M"));
I used clone because add modifies the original object, which might not be desired.

Answer 4

strtotime( "+1 month", strtotime( $time ) );
this returns a timestamp that can be used with the date function

Answer 5

You can use DateTime::modify like this :
$date = new DateTime('2010-12-11');
$date->modify('+1 month');
See documentations :
https://php.net/manual/en/datetime.modify.php
https://php.net/manual/en/class.datetime.php
UPDATE january 2021 : correct mistakes raised by comments
This solution has some problems for months with 31 days like May etc.
Exemple : this jumps from 31st May to 1st July which is incorrect.
To correct that, you can create this custom function
function addMonths($date,$months){
$init=clone $date;
$modifier=$months.' months';
$back_modifier =-$months.' months';
$date->modify($modifier);
$back_to_init= clone $date;
$back_to_init->modify($back_modifier);
while($init->format('m')!=$back_to_init->format('m')){
$date->modify('-1 day') ;
$back_to_init= clone $date;
$back_to_init->modify($back_modifier);
}
}
Then you can use it like that :
$date = new DateTime('2010-05-31');
addMonths($date, 1);
print_r($date);
//DateTime Object ( [date] => 2010-06-30 00:00:00.000000 [timezone_type] => 3 [timezone] => Europe/Berlin )
This solution was found in PHP.net posted by jenspj : https://www.php.net/manual/fr/datetime.modify.php#107592

Answer 6

(date('d') > 28) ? date("mdY", strtotime("last day of next month")) : date("mdY", strtotime("+1 month"));
This will compensate for February and the other 31 day months. You could of course do a lot more checking to to get more exact for 'this day next month' relative date formats (which does not work sadly, see below), and you could just as well use DateTime.
Both DateInterval('P1M') and strtotime("+1 month") are essentially blindly adding 31 days regardless of the number of days in the following month.
2010-01-31 => March 3rd
2012-01-31 => March 2nd (leap year)

Answer 7

Please first you set your date format as like 12-12-2012
After use this function it's work properly;
$date = date('d-m-Y',strtotime("12-12-2012 +2 Months");
Here 12-12-2012 is your date and +2 Months is increment of the month;
You also increment of Year, Date
strtotime("12-12-2012 +1 Year");
Ans is 12-12-2013

Answer 8

I use this way:-
$occDate='2014-01-28';
$forOdNextMonth= date('m', strtotime("+1 month", strtotime($occDate)));
//Output:- $forOdNextMonth=02
/*****************more example****************/
$occDate='2014-12-28';
$forOdNextMonth= date('m', strtotime("+1 month", strtotime($occDate)));
//Output:- $forOdNextMonth=01
//***********************wrong way**********************************//
$forOdNextMonth= date('m', strtotime("+1 month", $occDate));
//Output:- $forOdNextMonth=02; //instead of $forOdNextMonth=01;
//******************************************************************//

Answer 9

Just updating the answer with simple method for find the date after no of months. As the best answer marked doesn't give the correct solution.
<?php
$date = date('2020-05-31');
$current = date("m",strtotime($date));
$next = date("m",strtotime($date."+1 month"));
if($current==$next-1){
$needed = date('Y-m-d',strtotime($date." +1 month"));
}else{
$needed = date('Y-m-d', strtotime("last day of next month",strtotime($date)));
}
echo "Date after 1 month from 2020-05-31 would be : $needed";
?>

Answer 10

If you want to get the date of one month from now you can do it like this
echo date('Y-m-d', strtotime('1 month'));
If you want to get the date of two months from now, you can achieve that by doing this
echo date('Y-m-d', strtotime('2 month'));
And so on, that's all.

Answer 11

Thanks Jason, your post was very helpful. I reformatted it and added more comments to help me understand it all. In case that helps anyone, I have posted it here:
function cycle_end_date($cycle_start_date, $months) {
$cycle_start_date_object = new DateTime($cycle_start_date);
//Find the date interval that we will need to add to the start date
$date_interval = find_date_interval($months, $cycle_start_date_object);
//Add this date interval to the current date (the DateTime class handles remaining complexity like year-ends)
$cycle_end_date_object = $cycle_start_date_object->add($date_interval);
//Subtract (sub) 1 day from date
$cycle_end_date_object->sub(new DateInterval('P1D'));
//Format final date to Y-m-d
$cycle_end_date = $cycle_end_date_object->format('Y-m-d');
return $cycle_end_date;
}
//Find the date interval we need to add to start date to get end date
function find_date_interval($n_months, DateTime $cycle_start_date_object) {
//Create new datetime object identical to inputted one
$date_of_last_day_next_month = new DateTime($cycle_start_date_object->format('Y-m-d'));
//And modify it so it is the date of the last day of the next month
$date_of_last_day_next_month->modify('last day of +'.$n_months.' month');
//If the day of inputted date (e.g. 31) is greater than last day of next month (e.g. 28)
if($cycle_start_date_object->format('d') > $date_of_last_day_next_month->format('d')) {
//Return a DateInterval object equal to the number of days difference
return $cycle_start_date_object->diff($date_of_last_day_next_month);
//Otherwise the date is easy and we can just add a month to it
} else {
//Return a DateInterval object equal to a period (P) of 1 month (M)
return new DateInterval('P'.$n_months.'M');
}
}
$cycle_start_date = '2014-01-31'; // select date in Y-m-d format
$n_months = 1; // choose how many months you want to move ahead
$cycle_end_date = cycle_end_date($cycle_start_date, $n_months); // output: 2014-07-02

Answer 12

function dayOfWeek($date){
return DateTime::createFromFormat('Y-m-d', $date)->format('N');
}
Usage examples:
echo dayOfWeek(2016-12-22);
// "4"
echo dayOfWeek(date('Y-m-d'));
// "4"

Answer 13

$date = strtotime("2017-12-11");
$newDate = date("Y-m-d", strtotime("+1 month", $date));
If you want to increment by days you can also do it
$date = strtotime("2017-12-11");
$newDate = date("Y-m-d", strtotime("+5 day", $date));

Answer 14

For anyone looking for an answer to any date format.
echo date_create_from_format('d/m/Y', '15/04/2017')->add(new DateInterval('P1M'))->format('d/m/Y');
Just change the date format.

Answer 15

//ECHO MONTHS BETWEEN TWO TIMESTAMPS
$my_earliest_timestamp = 1532095200;
$my_latest_timestamp = 1554991200;
echo '<pre>';
echo "Earliest timestamp: ". date('c',$my_earliest_timestamp) ."\r\n";
echo "Latest timestamp: " .date('c',$my_latest_timestamp) ."\r\n\r\n";
echo "Month start of earliest timestamp: ". date('c',strtotime('first day of '. date('F Y',$my_earliest_timestamp))) ."\r\n";
echo "Month start of latest timestamp: " .date('c',strtotime('first day of '. date('F Y',$my_latest_timestamp))) ."\r\n\r\n";
echo "Month end of earliest timestamp: ". date('c',strtotime('last day of '. date('F Y',$my_earliest_timestamp)) + 86399) ."\r\n";
echo "Month end of latest timestamp: " .date('c',strtotime('last day of '. date('F Y',$my_latest_timestamp)) + 86399) ."\r\n\r\n";
$sMonth = strtotime('first day of '. date('F Y',$my_earliest_timestamp));
$eMonth = strtotime('last day of '. date('F Y',$my_earliest_timestamp)) + 86399;
$xMonth = strtotime('+1 month', strtotime('first day of '. date('F Y',$my_latest_timestamp)));
while ($eMonth < $xMonth) {
echo "Things from ". date('Y-m-d',$sMonth) ." to ". date('Y-m-d',$eMonth) ."\r\n\r\n";
$sMonth = $eMonth + 1; //add 1 second to bring forward last date into first second of next month.
$eMonth = strtotime('last day of '. date('F Y',$sMonth)) + 86399;
}

Answer 16

I find the mtkime() function works really well for this:
$start_date="2021-10-01";
$start_date_plus_a_month=date("Y-m-d", mktime(0, 0, 0, date("m",strtotime($start_date))+1, date("d",strtotime($start_date)), date("Y",strtotime($start_date))));
result: 2021-11-01
I like to subtract 1 from the 'day' to produce '2021-10-31' which can be useful if you want to display a range across 12 months, e.g. Oct 1, 2021 to Sep 30 2022
$start_date_plus_a_year=date("Y-m-d", mktime(0, 0, 0, date("m",strtotime($start_date))+12, date("d",strtotime($start_date))-1, date("Y",strtotime($start_date))));
result: 2022-09-30

Answer 17

The correct answer to the exact question asked is Giuseppe Canale's answer from earlier. I'm going to answer a slightly more generic question of how to increment the date by an arbitrary number of months, however.
<?php
/**
* Will return a timestamp corresponding to first day of the month that is N months into the future.
* #param int $months_later number of months into the future: 0 for current one
* #param string $today if supplied will be used as the "now" time
* #return int
*/
function rel_month_to_time($months_later, $today=null) {
if ($months_later===0) {
return is_null($today) ? time() : strtotime($today);
}
return strtotime('first day of next month', rel_month_to_time($months_later-1, $today));
}
As is many times the case, you can use recursion for these "human problems" like calendars. The above can be used to return a timestamp corresponding to "next month" -- the way we humans think of it.
<?php echo date('Y-m-d', rel_month_to_time(1, '2023-01-30'));
// 2023-02-01

Answer 18

As pointed by #NetVicious i corrected the code, it should work with all dates, some example:
2013-01-30 will be 2013-02-28
2013-05-15 will be 2013-05-15
2013-05-31 will be 2013-06-30
This code uses the DateTime class to create a new date object, then it adds 1 month to the date using the modify method. Next, it gets the day of the next month using the format method. If the next month's day doesn't match the original day, it modifies the date to the last day of the previous month using the modify method.
$original_date = "2013-01-30";
$original_day = date("d", strtotime($original_date));
$date = new DateTime($original_date);
$date->modify('+1 month');
$next_month_day = $date->format('d');
if ($next_month_day != $original_day) {
$date->modify('last day of previous month');
}
$new_date = $date->format('Y-m-d');
echo $new_date;

Answer 19

All presented solutions are not working properly.
strtotime() and DateTime::add or DateTime::modify give sometime invalid results.
Examples:
- 31.08.2019 + 1 month gives 01.10.2019 instead 30.09.2019
- 29.02.2020 + 1 year gives 01.03.2021 instead 28.02.2021
(tested on PHP 5.5, PHP 7.3)
Below is my function based on idea posted by Angelo that solves the problem:
// $time - unix time or date in any format accepted by strtotime() e.g. 2020-02-29
// $days, $months, $years - values to add
// returns new date in format 2021-02-28
function addTime($time, $days, $months, $years)
{
// Convert unix time to date format
if (is_numeric($time))
$time = date('Y-m-d', $time);
try
{
$date_time = new DateTime($time);
}
catch (Exception $e)
{
echo $e->getMessage();
exit;
}
if ($days)
$date_time->add(new DateInterval('P'.$days.'D'));
// Preserve day number
if ($months or $years)
$old_day = $date_time->format('d');
if ($months)
$date_time->add(new DateInterval('P'.$months.'M'));
if ($years)
$date_time->add(new DateInterval('P'.$years.'Y'));
// Patch for adding months or years
if ($months or $years)
{
$new_day = $date_time->format("d");
// The day is changed - set the last day of the previous month
if ($old_day != $new_day)
$date_time->sub(new DateInterval('P'.$new_day.'D'));
}
// You can chage returned format here
return $date_time->format('Y-m-d');
}
Usage examples:
echo addTime('2020-02-29', 0, 0, 1); // add 1 year (result: 2021-02-28)
echo addTime('2019-08-31', 0, 1, 0); // add 1 month (result: 2019-09-30)
echo addTime('2019-03-15', 12, 2, 1); // add 12 days, 2 months, 1 year (result: 2019-09-30)

Answer 20

put a date in input box then click the button get day from date in jquery
$(document).ready( function() {
$("button").click(function(){
var day = ["Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"];
var a = new Date();
$(".result").text(day[a.getDay()]);
});
});

Answer 21

<?php
$selectdata ="select fromd,tod from register where username='$username'";
$q=mysqli_query($conm,$selectdata);
$row=mysqli_fetch_array($q);
$startdate=$row['fromd'];
$stdate=date('Y', strtotime($startdate));
$endate=$row['tod'];
$enddate=date('Y', strtotime($endate));
$years = range ($stdate,$enddate);
echo '<select name="years" class="form-control">';
echo '<option>SELECT</option>';
foreach($years as $year)
{ echo '<option value="'.$year.'"> '.$year.' </option>'; }
echo '</select>'; ?>