用未来最大的1减去过去最大的1，然后除以60。

时间是Unix格式的，所以它们只是一个大数字，显示了从格林尼治时间1970年1月1日00:00:00开始的秒数

<?php
$start = strtotime('12:01:00');
$end = strtotime('13:16:00');
$mins = ($end - $start) / 60;
echo $mins;
?>

输出:

75

时区的另一种方法。

$start_date = new DateTime("2013-12-24 06:00:00",new DateTimeZone('Pacific/Nauru'));
$end_date = new DateTime("2013-12-24 06:45:00", new DateTimeZone('Pacific/Nauru'));
$interval = $start_date->diff($end_date);
$hours   = $interval->format('%h'); 
$minutes = $interval->format('%i');
echo  'Diff. in minutes is: '.($hours * 60 + $minutes);

它在我的程序上工作，我使用date_diff，你可以在这里检查date_diff手册。

$start = date_create('2015-01-26 12:01:00');
$end = date_create('2015-01-26 13:15:00');
$diff=date_diff($end,$start);
print_r($diff);

你得到你想要的结果。

这就是我如何显示“xx次前”在php > 5.2 ..这里是DateTime对象的更多信息

//Usage:
$pubDate = $row['rssfeed']['pubDates']; // e.g. this could be like 'Sun, 10 Nov 2013 14:26:00 GMT'
$diff = ago($pubDate);    // output: 23 hrs ago

// Return the value of time different in "xx times ago" format
function ago($timestamp)
{

    $today = new DateTime(date('y-m-d h:i:s')); // [2]
    //$thatDay = new DateTime('Sun, 10 Nov 2013 14:26:00 GMT');
    $thatDay = new DateTime($timestamp);
    $dt = $today->diff($thatDay);

    if ($dt->y > 0){
        $number = $dt->y;
        $unit = "year";
    } else if ($dt->m > 0) {
        $number = $dt->m;
        $unit = "month";
    } else if ($dt->d > 0) {
        $number = $dt->d;
        $unit = "day";
    } else if ($dt->h > 0) {
        $number = $dt->h;
        $unit = "hour";
    } else if ($dt->i > 0) {
        $number = $dt->i;
        $unit = "minute";
    } else if ($dt->s > 0) {
        $number = $dt->s;
        $unit = "second";
    }
    
    $unit .= $number  > 1 ? "s" : "";
 
    $ret = $number." ".$unit." "."ago";
    return $ret;
}

$date1=date_create("2020-03-15");
$date2=date_create("2020-12-12");
$diff=date_diff($date1,$date2);
echo $diff->format("%R%a days");

有关详细的格式说明，请访问链接。