PHP DateTime::modify adding and subtracting months

问题:

I've been working a lot with the DateTime class and recently ran into what I thought was a bug when adding months. After a bit of research, it appears that it wasn't a bug, but instead working as intended. According to the documentation found here:

Example #2 Beware when adding or subtracting months

<?php
$date = new DateTime('2000-12-31');

$date->modify('+1 month');
echo $date->format('Y-m-d') . "\n";

$date->modify('+1 month');
echo $date->format('Y-m-d') . "\n";
?>

The above example will output:
2001-01-31
2001-03-03

Can anyone justify why this isn't considered a bug?

Furthermore does anyone have any elegant solutions to correct the issue and make it so +1 month will work as expected instead of as intended?

回答1:

Why it's not a bug:

The current behavior is correct. The following happens internally:

1. +1 month increases the month number (originally 1) by one. This makes the date 2010-02-31.

2. The second month (February) only has 28 days in 2010, so PHP auto-corrects this by just continuing to count days from February 1st. You then end up at March 3rd.

How to get what you want:

To get what you want is by: manually checking the next month. Then add the number of days next month has.

I hope you can yourself code this. I am just giving what-to-do.

PHP 5.3 way:

To obtain the correct behavior, you can use one of the PHP 5.3's new functionality that introduces the relative time stanza first day of. This stanza can be used in combination with next month, fifth month or +8 months to go to the first day of the specified month. Instead of +1 month from what you're doing, you can use this code to get the first day of next month like this:

<?php
$d = new DateTime( '2010-01-31' );
$d->modify( 'first day of next month' );
echo $d->format( 'F' ), "\n";
?>

This script will correctly output February. The following things happen when PHP processes this first day of next month stanza:

1. next month increases the month number (originally 1) by one. This makes the date 2010-02-31.

2. first day of sets the day number to 1, resulting in the date 2010-02-01.

回答2:

This may be useful:

echo Date("Y-m-d", strtotime("2013-01-01 +1 Month -1 Day"));
  // 2013-01-31

echo Date("Y-m-d", strtotime("2013-02-01 +1 Month -1 Day"));
  // 2013-02-28

echo Date("Y-m-d", strtotime("2013-03-01 +1 Month -1 Day"));
  // 2013-03-31

echo Date("Y-m-d", strtotime("2013-04-01 +1 Month -1 Day"));
  // 2013-04-30

echo Date("Y-m-d", strtotime("2013-05-01 +1 Month -1 Day"));
  // 2013-05-31

echo Date("Y-m-d", strtotime("2013-06-01 +1 Month -1 Day"));
  // 2013-06-30

echo Date("Y-m-d", strtotime("2013-07-01 +1 Month -1 Day"));
  // 2013-07-31

echo Date("Y-m-d", strtotime("2013-08-01 +1 Month -1 Day"));
  // 2013-08-31

echo Date("Y-m-d", strtotime("2013-09-01 +1 Month -1 Day"));
  // 2013-09-30

echo Date("Y-m-d", strtotime("2013-10-01 +1 Month -1 Day"));
  // 2013-10-31

echo Date("Y-m-d", strtotime("2013-11-01 +1 Month -1 Day"));
  // 2013-11-30

echo Date("Y-m-d", strtotime("2013-12-01 +1 Month -1 Day"));
  // 2013-12-31

回答3:

My solution to the problem:

$startDate = new \DateTime( '2015-08-30' );
$endDate = clone $startDate;

$billing_count = '6';
$billing_unit = 'm';

$endDate->add( new \DateInterval( 'P' . $billing_count . strtoupper( $billing_unit ) ) );

if ( intval( $endDate->format( 'n' ) ) > ( intval( $startDate->format( 'n' ) ) + intval( $billing_count ) ) % 12 )
{
    if ( intval( $startDate->format( 'n' ) ) + intval( $billing_count ) != 12 )
    {
        $endDate->modify( 'last day of -1 month' );
    }
}

回答4:

Here is another compact solution entirely using DateTime methods, modifying the object in-place without creating clones.

$dt = new DateTime('2012-01-31');

echo $dt->format('Y-m-d'), PHP_EOL;

$day = $dt->format('j');
$dt->modify('first day of +1 month');
$dt->modify('+' . (min($day, $dt->format('t')) - 1) . ' days');

echo $dt->format('Y-m-d'), PHP_EOL;

It outputs:

2012-01-31
2012-02-29

回答5:

I made a function that returns a DateInterval to make sure that adding a month shows the next month, and removes the days into the after that.

$time = new DateTime('2014-01-31');
echo $time->format('d-m-Y H:i') . '<br/>';

$time->add( add_months(1, $time));

echo $time->format('d-m-Y H:i') . '<br/>';



function add_months( $months, \DateTime $object ) {
    $next = new DateTime($object->format('d-m-Y H:i:s'));
    $next->modify('last day of +'.$months.' month');

    if( $object->format('d') > $next->format('d') ) {
        return $object->diff($next);
    } else {
        return new DateInterval('P'.$months.'M');
    }
}

回答6:

In conjunction with shamittomar's answer, it could then be this for adding months "safely":

/**
 * Adds months without jumping over last days of months
 *
 * @param \DateTime $date
 * @param int $monthsToAdd
 * @return \DateTime
 */

public function addMonths($date, $monthsToAdd) {
    $tmpDate = clone $date;
    $tmpDate->modify('first day of +'.(int) $monthsToAdd.' month');

    if($date->format('j') > $tmpDate->format('t')) {
        $daysToAdd = $tmpDate->format('t') - 1;
    }else{
        $daysToAdd = $date->format('j') - 1;
    }

    $tmpDate->modify('+ '. $daysToAdd .' days');


    return $tmpDate;
}

回答7:

I agree with the sentiment of the OP that this is counter-intuitive and frustrating, but so is determining what +1 month means in the scenarios where this occurs. Consider these examples:

You start with 2015-01-31 and want to add a month 6 times to get a scheduling cycle for sending an email newsletter. With the OP's initial expectations in mind, this would return:

• 2015-01-31
• 2015-02-28
• 2015-03-31
• 2015-04-30
• 2015-05-31
• 2015-06-30

Right away, notice that we are expecting +1 month to mean last day of month or, alternatively, to add 1 month per iteration but always in reference to the start point. Instead of interpreting this as "last day of month" we could read it as "31st day of next month or last available within that month". This means that we jump from April 30th to May 31st instead of to May 30th. Note that this is not because it is "last day of month" but because we want "closest available to date of start month."

So suppose one of our users subscribes to another newsletter to start on 2015-01-30. What is the intuitive date for +1 month? One interpretation would be "30th day of next month or closest available" which would return:

• 2015-01-30
• 2015-02-28
• 2015-03-30
• 2015-04-30
• 2015-05-30
• 2015-06-30

This would be fine except when our user gets both newsletters on the same day. Let's assume that this is a supply-side issue instead of demand-side We're not worried that the user will be annoyed with getting 2 newsletters in the same day but instead that our mail servers can't afford the bandwidth for sending twice as many newsletters. With that in mind, we return to the other interpretation of "+1 month" as "send on the second to last day of each month" which would return:

• 2015-01-30
• 2015-02-27
• 2015-03-30
• 2015-04-29
• 2015-05-30
• 2015-06-29

Now we've avoided any overlap with the first set, but we also end up with April and June 29th, which certainly does match our original intuitions that +1 month simply should return m/$d/Y or the attractive and simple m/30/Y for all possible months. So now let's consider a third interpretation of +1 month using both dates:

Jan. 31st

• 2015-01-31
• 2015-03-03
• 2015-03-31
• 2015-05-01
• 2015-05-31
• 2015-07-01

Jan. 30th

• 2015-01-30
• 2015-03-02
• 2015-03-30
• 2015-04-30
• 2015-05-30
• 2015-06-30

The above has some issues. February is skipped, which could be a problem both supply-end (say if there is a monthly bandwidth allocation and Feb goes to waste and March gets doubled up on) and demand-end (users feel cheated out of Feb and perceive the extra March as attempt to correct mistake). On the other hand, notice that the two date sets:

• never overlap
• are always on the same date when that month has the date (so the Jan. 30 set looks pretty clean)
• are all within 3 days (1 day in most cases) of what might be considered the "correct" date.
• are all at least 28 days (a lunar month) from their successor and predecessor, so very evenly distributed.

Given the last two sets, it would not be difficult to simply roll back one of the dates if it falls outside of the actual following month (so roll back to Feb 28th and April 30th in the first set) and not lose any sleep over the occasional overlap and divergence from the "last day of month" vs "second to last day of month" pattern. But expecting the library to choose between "most pretty/natural", "mathematical interpretation of 02/31 and other month overflows", and "relative to first of month or last month" is always going to end with someone's expectations not being met and some schedule needing to adjust the "wrong" date to avoid the real-world problem that the "wrong" interpretation introduces.

So again, while I also would expect +1 month to return a date that actually is in the following month, it is not as simple as intuition and given the choices, going with math over the expectations of web developers is probably the safe choice.

Here's an alternative solution that is still as clunky as any but I think has nice results:

foreach(range(0,5) as $count) {
    $new_date = clone $date;
    $new_date->modify("+$count month");
    $expected_month = $count + 1;
    $actual_month = $new_date->format("m");
    if($expected_month != $actual_month) {
        $new_date = clone $date;
        $new_date->modify("+". ($count - 1) . " month");
        $new_date->modify("+4 weeks");
    }

    echo "* " . nl2br($new_date->format("Y-m-d") . PHP_EOL);
}

It's not optimal but the underlying logic is : If adding 1 month results in a date other than the expected next month, scrap that date and add 4 weeks instead. Here are the results with the two test dates:

Jan. 31st

• 2015-01-31
• 2015-02-28
• 2015-03-31
• 2015-04-28
• 2015-05-31
• 2015-06-28

Jan. 30th

• 2015-01-30
• 2015-02-27
• 2015-03-30
• 2015-04-30
• 2015-05-30
• 2015-06-30

(My code is a mess and wouldn't work in a multi-year scenario. I welcome anyone to rewrite the solution with more elegant code so long as the underlying premise is kept intact, i.e. if +1 month returns a funky date, use +4 weeks instead.)

回答8:

Here is an implementation of an improved version of Juhana's answer in a related question:

<?php
function sameDateNextMonth(DateTime $createdDate, DateTime $currentDate) {
    $addMon = clone $currentDate;
    $addMon->add(new DateInterval("P1M"));

    $nextMon = clone $currentDate;
    $nextMon->modify("last day of next month");

    if ($addMon->format("n") == $nextMon->format("n")) {
        $recurDay = $createdDate->format("j");
        $daysInMon = $addMon->format("t");
        $currentDay = $currentDate->format("j");
        if ($recurDay > $currentDay && $recurDay <= $daysInMon) {
            $addMon->setDate($addMon->format("Y"), $addMon->format("n"), $recurDay);
        }
        return $addMon;
    } else {
        return $nextMon;
    }
}

This version takes $createdDate under the presumption that you are dealing with a recurring monthly period, such as a subscription, that started on a specific date, such as the 31st. It always takes $createdDate so late "recurs on" dates won't shift to lower values as they are pushed forward thru lesser-valued months (e.g., so all 29th, 30th or 31st recur dates won't eventually get stuck on the 28th after passing thru a non-leap-year February).

Here is some driver code to test the algorithm:

$createdDate = new DateTime("2015-03-31");
echo "created date = " . $createdDate->format("Y-m-d") . PHP_EOL;

$next = sameDateNextMonth($createdDate, $createdDate);
echo "   next date = " . $next->format("Y-m-d") . PHP_EOL;

foreach(range(1, 12) as $i) {
    $next = sameDateNextMonth($createdDate, $next);
    echo "   next date = " . $next->format("Y-m-d") . PHP_EOL;
}

Which outputs:

created date = 2015-03-31
   next date = 2015-04-30
   next date = 2015-05-31
   next date = 2015-06-30
   next date = 2015-07-31
   next date = 2015-08-31
   next date = 2015-09-30
   next date = 2015-10-31
   next date = 2015-11-30
   next date = 2015-12-31
   next date = 2016-01-31
   next date = 2016-02-29
   next date = 2016-03-31
   next date = 2016-04-30

回答9:

I found a shorter way around it using the following code:

                   $datetime = new DateTime("2014-01-31");
                    $month = $datetime->format('n'); //without zeroes
                    $day = $datetime->format('j'); //without zeroes

                    if($day == 31){
                        $datetime->modify('last day of next month');
                    }else if($day == 29 || $day == 30){
                        if($month == 1){
                            $datetime->modify('last day of next month');                                
                        }else{
                            $datetime->modify('+1 month');                                
                        }
                    }else{
                        $datetime->modify('+1 month');
                    }
echo $datetime->format('Y-m-d H:i:s');

回答10:

This is an improved version of Kasihasi's answer in a related question. This will correctly add or subtract an arbitrary number of months to a date.

public static function addMonths($monthToAdd, $date) {
    $d1 = new DateTime($date);

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

    if ($monthToAdd > 0) {
        $year += floor($monthToAdd/12);
    } else {
        $year += ceil($monthToAdd/12);
    }
    $monthToAdd = $monthToAdd%12;
    $month += $monthToAdd;
    if($month > 12) {
        $year ++;
        $month -= 12;
    } elseif ($month < 1 ) {
        $year --;
        $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);
    }
    return $d2->format('Y-m-d');
}

For example:

addMonths(-25, '2017-03-31')

will output:

'2015-02-28'

回答11:

If you just want to avoid skipping a month you can perform something like this to get the date out and run a loop on the next month reducing the date by one and rechecking until a valid date where $starting_calculated is a valid string for strtotime (i.e. mysql datetime or "now"). This finds the very end of the month at 1 minute to midnight instead of skipping the month.

    $start_dt = $starting_calculated;

    $next_month = date("m",strtotime("+1 month",strtotime($start_dt)));
    $next_month_year = date("Y",strtotime("+1 month",strtotime($start_dt)));

    $date_of_month = date("d",$starting_calculated);

    if($date_of_month>28){
        $check_date = false;
        while(!$check_date){
            $check_date = checkdate($next_month,$date_of_month,$next_month_year);
            $date_of_month--;
        }
        $date_of_month++;
        $next_d = $date_of_month;
    }else{
        $next_d = "d";
    }
    $end_dt = date("Y-m-$next_d 23:59:59",strtotime("+1 month"));

回答12:

Extension for DateTime class which solves problem of adding or subtracting months

https://gist.github.com/66Ton99/60571ee49bf1906aaa1c

回答13:

If using strtotime() just use $date = strtotime('first day of +1 month');

回答14:

you can actually do it with just date() and strtotime() as well. For example to add 1 month to todays date:

date("Y-m-d",strtotime("+1 month",time()));

if you are wanting to use the datetime class thats fine too but this is just as easy. more details here

回答15:

I needed to get a date for 'this month last year' and it becomes unpleasant quite quickly when this month is February in a leap year. However, I believe this works... :-/ The trick seems to be to base your change on the 1st day of the month.

$this_month_last_year_end = new \DateTime();
$this_month_last_year_end->modify('first day of this month');
$this_month_last_year_end->modify('-1 year');
$this_month_last_year_end->modify('last day of this month');
$this_month_last_year_end->setTime(23, 59, 59);

回答16:

$ds = new DateTime();
$ds->modify('+1 month');
$ds->modify('first day of this month');

回答17:

$month = 1; $year = 2017;
echo date('n', mktime(0, 0, 0, $month + 2, -1, $year));

will output 2 (february). will work for other months too.

回答18:

     $date = date('Y-m-d', strtotime("+1 month"));
     echo $date;