如何在PHP中计算两个日期时间之间的分钟差异?


当前回答

求两个日期之差的方法在这里。有了这个功能,你可以找到秒、分、小时、天、年和月等差异。

function alihan_diff_dates($date = null, $diff = "minutes") {
 $start_date = new DateTime($date);
 $since_start = $start_date->diff(new DateTime( date('Y-m-d H:i:s') )); // date now
 print_r($since_start);
 switch ($diff) {
    case 'seconds':
        return $since_start->s;
        break;
    case 'minutes':
        return $since_start->i;
        break;
    case 'hours':
        return $since_start->h;
        break;
    case 'days':
        return $since_start->d;
        break;      
    default:
        # code...
        break;
 }
}

你可以开发这个函数。我测试过了,结果对我有用。DateInterval对象输出如下:

/*
DateInterval Object ( [y] => 0 [m] => 0 [d] => 0 [h] => 0 [i] => 5 [s] => 13 [f] => 0 [weekday] => 0 [weekday_behavior] => 0 [first_last_day_of] => 0 [invert] => 0 [days] => 0 [special_type] => 0 [special_amount] => 0 [have_weekday_relative] => 0 [have_special_relative] => 0 ) 
*/

功能用途:

$date =过去的日期, $diff =类型eg:“分钟”,“天”,“秒”

$diff_mins = alihan_diff_dates("2019-03-24 13:24:19", "minutes");

祝你好运。

其他回答

以上答案适用于较旧版本的PHP。使用DateTime类来进行任何日期计算,因为PHP 5.3是标准的。 如。

$start_date = new DateTime('2007-09-01 04:10:58');
$since_start = $start_date->diff(new DateTime('2012-09-11 10:25:00'));
echo $since_start->days.' days total<br>';
echo $since_start->y.' years<br>';
echo $since_start->m.' months<br>';
echo $since_start->d.' days<br>';
echo $since_start->h.' hours<br>';
echo $since_start->i.' minutes<br>';
echo $since_start->s.' seconds<br>';

$since_start是一个DateInterval对象。注意,days属性是可用的(因为我们使用了DateTime类的diff方法来生成DateInterval对象)。

上面的代码将输出:

1837天总共5年0个月10天6小时14分2秒

获取总分钟数:

$minutes = $since_start->days * 24 * 60;
$minutes += $since_start->h * 60;
$minutes += $since_start->i;
echo $minutes.' minutes';

这将输出:

2645654分钟

也就是这两个日期之间的实际分钟数。DateTime类将考虑夏令时(取决于时区),而“旧方法”不会考虑。阅读有关日期和时间的手册http://www.php.net/manual/en/book.datetime.php

这是一个简单的一行代码:

$start = new DateTime('yesterday');
$end = new DateTime('now');
$diffInMinutes = iterator_count(new \DatePeriod($start, new \DateInterval('PT1M'), $end));

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

求两个日期之差的方法在这里。有了这个功能,你可以找到秒、分、小时、天、年和月等差异。

function alihan_diff_dates($date = null, $diff = "minutes") {
 $start_date = new DateTime($date);
 $since_start = $start_date->diff(new DateTime( date('Y-m-d H:i:s') )); // date now
 print_r($since_start);
 switch ($diff) {
    case 'seconds':
        return $since_start->s;
        break;
    case 'minutes':
        return $since_start->i;
        break;
    case 'hours':
        return $since_start->h;
        break;
    case 'days':
        return $since_start->d;
        break;      
    default:
        # code...
        break;
 }
}

你可以开发这个函数。我测试过了,结果对我有用。DateInterval对象输出如下:

/*
DateInterval Object ( [y] => 0 [m] => 0 [d] => 0 [h] => 0 [i] => 5 [s] => 13 [f] => 0 [weekday] => 0 [weekday_behavior] => 0 [first_last_day_of] => 0 [invert] => 0 [days] => 0 [special_type] => 0 [special_amount] => 0 [have_weekday_relative] => 0 [have_special_relative] => 0 ) 
*/

功能用途:

$date =过去的日期, $diff =类型eg:“分钟”,“天”,“秒”

$diff_mins = alihan_diff_dates("2019-03-24 13:24:19", "minutes");

祝你好运。

function date_getFullTimeDifference( $start, $end )
{
$uts['start']      =    strtotime( $start );
        $uts['end']        =    strtotime( $end );
        if( $uts['start']!==-1 && $uts['end']!==-1 )
        {
            if( $uts['end'] >= $uts['start'] )
            {
                $diff    =    $uts['end'] - $uts['start'];
                if( $years=intval((floor($diff/31104000))) )
                    $diff = $diff % 31104000;
                if( $months=intval((floor($diff/2592000))) )
                    $diff = $diff % 2592000;
                if( $days=intval((floor($diff/86400))) )
                    $diff = $diff % 86400;
                if( $hours=intval((floor($diff/3600))) )
                    $diff = $diff % 3600;
                if( $minutes=intval((floor($diff/60))) )
                    $diff = $diff % 60;
                $diff    =    intval( $diff );
                return( array('years'=>$years,'months'=>$months,'days'=>$days, 'hours'=>$hours, 'minutes'=>$minutes, 'seconds'=>$diff) );
            }
            else
            {
                echo "Ending date/time is earlier than the start date/time";
            }
        }
        else
        {
            echo "Invalid date/time data detected";
        }
}