在 PHP 中，如何获得两个时间戳之间准确（而非近似）的周数、月数和年数？

Question

I can do it for days like this:

$d1 = new DateTime('2000-01-01 12:00:00');
$d2 = new DateTime('2020-01-01 12:00:00');
$diff = $d2->diff($d1);
echo $diff->days;

In other words, it works for days . However, the DateTime/ DateInterval class has only a $days variable -- these are expected but don't exist :

$diff->weeks;
$diff->months;
$diff->years;

Reading the manual, you might at first glance be deceived into thinking that it does have these attributes: https://www.php.net/manual/en/class.dateinterval.php

public integer $y ;
public integer $m ;
public integer $d ;
public integer $h ;
public integer $i ;
public integer $s ;
public float $f ;
public integer $invert ;
public mixed $days ;

The y, m, d, h, i, s there are not "individual totals", but depend on each other. For example, if the time span is exactly one year, the $y will be 1, but all of the other ones will be 0, instead of their respective representations (12 months, 52 weeks, etc.).

They treat days specially for some reason by including the $days variable, which does show the actual total number of days. I want that for weeks, months and years too.

I already know how to "estimate" the number of weeks/months/years between two timestamps, by using simple math and fixed variables representing the average number of seconds in each time unit. Since this doesn't take into consideration all the complexities of "traversing" the calendar format(s), such as leap years, varying days in different months, and many other small/complex details, you don't get the exact number that way.

I want to know the exact total number of weeks between two timestamps, and the same thing for years and months, independent of each other.

Answer 1

This will return the exact difference between two days hope this will help you.

$time_diffrence=getDiffrenceBetweenTwoDays($date1,$date2); 

function getDiffrenceBetweenTwoDays($date1,$date2){
    $etime = strtotime($date1) - strtotime($date2;

    if ($etime < 1)
    {
        return '0 seconds';
    }

    $a = array( 365 * 24 * 60 * 60  =>  'year',
                 30 * 24 * 60 * 60  =>  'month',
                      24 * 60 * 60  =>  'day',
                           60 * 60  =>  'hour',
                                60  =>  'minute',
                                 1  =>  'second'
                );
    $a_plural = array( 'year'   => 'years',
                       'month'  => 'months',
                       'day'    => 'days',
                       'hour'   => 'hours',
                       'minute' => 'minutes',
                       'second' => 'seconds'
                );

    foreach ($a as $secs => $str)
    {
        $d = $etime / $secs;
        if ($d >= 1)
        {
            $r = round($d);
            return $r . ' ' . ($r > 1 ? $a_plural[$str] : $str) .''  ;
        }
    }
}

Answer 2

Replace %a with any of the following at this link:

FORMATS

$d1 = date_create('2000-01-01 12:00:00');
$d2 = date_create('2020-01-01 12:00:00');
$diff = date_diff($d1, $d2);

$days = $diff->format('%a');
echo $days; // 7305