由于每个人都在发布代码示例，这里有另一个版本。

我想要一个函数来显示从秒到年的差异（仅一个单位）。对于超过1天的时段，我希望它在午夜滚动（周一上午10点到周三上午9点是2天前，而不是1天前）。对于超过一个月的时间段，我希望滚动在当月的同一天（包括30/31天的月份和闰年）。

这就是我想到的：

/**
 * Returns how long ago something happened in the past, showing it
 * as n seconds / minutes / hours / days / weeks / months / years ago.
 *
 * For periods over a day, it rolls over at midnight (so doesn't depend
 * on current time of day), and it correctly accounts for month-lengths
 * and leap-years (months and years rollover on current day of month).
 *
 * $param string $timestamp in DateTime format
 * $return string description of interval
 */
function ago($timestamp)
{
    $then = date_create($timestamp);

    // for anything over 1 day, make it rollover on midnight
    $today = date_create('tomorrow'); // ie end of today
    $diff = date_diff($then, $today);

    if ($diff->y > 0) return $diff->y.' year'.($diff->y>1?'s':'').' ago';
    if ($diff->m > 0) return $diff->m.' month'.($diff->m>1?'s':'').' ago';
    $diffW = floor($diff->d / 7);
    if ($diffW > 0) return $diffW.' week'.($diffW>1?'s':'').' ago';
    if ($diff->d > 1) return $diff->d.' day'.($diff->d>1?'s':'').' ago';

    // for anything less than 1 day, base it off 'now'
    $now = date_create();
    $diff = date_diff($then, $now);

    if ($diff->d > 0) return 'yesterday';
    if ($diff->h > 0) return $diff->h.' hour'.($diff->h>1?'s':'').' ago';
    if ($diff->i > 0) return $diff->i.' minute'.($diff->i>1?'s':'').' ago';
    return $diff->s.' second'.($diff->s==1?'':'s').' ago';
}

查看小时、分钟和秒。。

$date1 = "2008-11-01 22:45:00"; 

$date2 = "2009-12-04 13:44:01"; 

$diff = abs(strtotime($date2) - strtotime($date1)); 

$years   = floor($diff / (365*60*60*24)); 
$months  = floor(($diff - $years * 365*60*60*24) / (30*60*60*24)); 
$days    = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24)/ (60*60*24));

$hours   = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24)/ (60*60)); 

$minuts  = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24 - $hours*60*60)/ 60); 

$seconds = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24 - $hours*60*60 - $minuts*60)); 

printf("%d years, %d months, %d days, %d hours, %d minuts\n, %d seconds\n", $years, $months, $days, $hours, $minuts, $seconds); 

这是我的职责。所需PHP>=5.3.4。它使用DateTime类。非常快，很快，可以区分两个日期，甚至所谓的“开始时间”。

if(function_exists('grk_Datetime_Since') === FALSE){
    function grk_Datetime_Since($From, $To='', $Prefix='', $Suffix=' ago', $Words=array()){
        #   Est-ce qu'on calcul jusqu'à un moment précis ? Probablement pas, on utilise maintenant
        if(empty($To) === TRUE){
            $To = time();
        }

        #   On va s'assurer que $From est numérique
        if(is_int($From) === FALSE){
            $From = strtotime($From);
        };

        #   On va s'assurer que $To est numérique
        if(is_int($To) === FALSE){
            $To = strtotime($To);
        }

        #   On a une erreur ?
        if($From === FALSE OR $From === -1 OR $To === FALSE OR $To === -1){
            return FALSE;
        }

        #   On va créer deux objets de date
        $From = new DateTime(@date('Y-m-d H:i:s', $From), new DateTimeZone('GMT'));
        $To   = new DateTime(@date('Y-m-d H:i:s', $To), new DateTimeZone('GMT'));

        #   On va calculer la différence entre $From et $To
        if(($Diff = $From->diff($To)) === FALSE){
            return FALSE;
        }

        #   On va merger le tableau des noms (par défaut, anglais)
        $Words = array_merge(array(
            'year'      => 'year',
            'years'     => 'years',
            'month'     => 'month',
            'months'    => 'months',
            'week'      => 'week',
            'weeks'     => 'weeks',
            'day'       => 'day',
            'days'      => 'days',
            'hour'      => 'hour',
            'hours'     => 'hours',
            'minute'    => 'minute',
            'minutes'   => 'minutes',
            'second'    => 'second',
            'seconds'   => 'seconds'
        ), $Words);

        #   On va créer la chaîne maintenant
        if($Diff->y > 1){
            $Text = $Diff->y.' '.$Words['years'];
        } elseif($Diff->y == 1){
            $Text = '1 '.$Words['year'];
        } elseif($Diff->m > 1){
            $Text = $Diff->m.' '.$Words['months'];
        } elseif($Diff->m == 1){
            $Text = '1 '.$Words['month'];
        } elseif($Diff->d > 7){
            $Text = ceil($Diff->d/7).' '.$Words['weeks'];
        } elseif($Diff->d == 7){
            $Text = '1 '.$Words['week'];
        } elseif($Diff->d > 1){
            $Text = $Diff->d.' '.$Words['days'];
        } elseif($Diff->d == 1){
            $Text = '1 '.$Words['day'];
        } elseif($Diff->h > 1){
            $Text = $Diff->h.' '.$Words['hours'];
        } elseif($Diff->h == 1){
            $Text = '1 '.$Words['hour'];
        } elseif($Diff->i > 1){
            $Text = $Diff->i.' '.$Words['minutes'];
        } elseif($Diff->i == 1){
            $Text = '1 '.$Words['minute'];
        } elseif($Diff->s > 1){
            $Text = $Diff->s.' '.$Words['seconds'];
        } else {
            $Text = '1 '.$Words['second'];
        }

        return $Prefix.$Text.$Suffix;
    }
}

您还可以使用以下代码通过向上舍入分数来返回日期差异$date1=$duedate；//指定到期日echo$date2=日期（“Y-m-d”）；//当前日期$ts1=字符串时间（$date1）；$ts2=字符串时间（$date2）；$seconds_diff=$ts1-$ts2；echo$datediff=ceil（（$seconds_diff/3600）/24）；//天内返回

如果您使用php的floor方法而不是ceil，它将返回舍入分数。请检查此处的差异，有时，如果您的临时服务器时区与现场站点时区不同，在这种情况下，您可能会得到不同的结果，因此请相应地更改条件。

一便士一英镑：我刚刚回顾了几个解决方案，所有这些方案都使用floor（）提供了一个复杂的解决方案，然后四舍五入到26年12个月零2天的解决方案中，原本应该是25年11个月零20天！！！！

这是我对这个问题的看法：可能不优雅，可能编码不好，但如果不计算LEAP年份，则提供了更接近答案的答案，显然闰年可以编码为，但在这种情况下-正如其他人所说，也许您可以提供以下答案：我已经包含了所有测试条件和print_r，以便您可以更清楚地看到结果的构造：在这里，

//设置输入日期/变量：：

$ISOstartDate   = "1987-06-22";
$ISOtodaysDate = "2013-06-22";

//我们需要将ISO yyyy-mm-dd格式分解为yyyy-mm-d格式，如下所示：

$yDate[]=爆炸（'-'，$ISOstartDate）；print_r（$yDate）；

$zDate[]=爆炸（'-'，$ISOtodaysDate）；print_r（$zDate）；

// Lets Sort of the Years!
// Lets Sort out the difference in YEARS between startDate and todaysDate ::
$years = $zDate[0][0] - $yDate[0][0];

// We need to collaborate if the month = month = 0, is before or after the Years Anniversary ie 11 months 22 days or 0 months 10 days...
if ($months == 0 and $zDate[0][1] > $ydate[0][1]) {
    $years = $years -1;
}
// TEST result
echo "\nCurrent years => ".$years;

// Lets Sort out the difference in MONTHS between startDate and todaysDate ::
$months = $zDate[0][1] - $yDate[0][1];

// TEST result
echo "\nCurrent months => ".$months;

// Now how many DAYS has there been - this assumes that there is NO LEAP years, so the calculation is APPROXIMATE not 100%
// Lets cross reference the startDates Month = how many days are there in each month IF m-m = 0 which is a years anniversary
// We will use a switch to check the number of days between each month so we can calculate days before and after the years anniversary

switch ($yDate[0][1]){
    case 01:    $monthDays = '31';  break;  // Jan
    case 02:    $monthDays = '28';  break;  // Feb
    case 03:    $monthDays = '31';  break;  // Mar
    case 04:    $monthDays = '30';  break;  // Apr
    case 05:    $monthDays = '31';  break;  // May
    case 06:    $monthDays = '30';  break;  // Jun
    case 07:    $monthDays = '31';  break;  // Jul
    case 08:    $monthDays = '31';  break;  // Aug
    case 09:    $monthDays = '30';  break;  // Sept
    case 10:    $monthDays = '31';  break;  // Oct
    case 11:    $monthDays = '30';  break;  // Nov
    case 12:    $monthDays = '31';  break;  // Dec
};
// TEST return
echo "\nDays in start month ".$yDate[0][1]." => ".$monthDays;


// Lets correct the problem with 0 Months - is it 11 months + days, or 0 months +days???

$days = $zDate[0][2] - $yDate[0][2] +$monthDays;
echo "\nCurrent days => ".$days."\n";

// Lets now Correct the months to being either 11 or 0 Months, depending upon being + or - the years Anniversary date 
// At the same time build in error correction for Anniversary dates not being 1yr 0m 31d... see if ($days == $monthDays )
if($days < $monthDays && $months == 0)
    {
    $months = 11;       // If Before the years anniversary date
    }
else    {
    $months = 0;        // If After the years anniversary date
    $years = $years+1;  // Add +1 to year
    $days = $days-$monthDays;   // Need to correct days to how many days after anniversary date
    };
// Day correction for Anniversary dates
if ($days == $monthDays )   // if todays date = the Anniversary DATE! set days to ZERO
    {
    $days = 0;          // days set toZERO so 1 years 0 months 0 days
    };

    echo "\nTherefore, the number of years/ months/ days/ \nbetween start and todays date::\n\n";

    printf("%d years, %d months, %d days\n", $years, $months, $days);

最终结果是：26年零个月零天

这就是我在2013年6月22日做生意的时间——哎呦！

使用此函数

//function Diff between Dates
//////////////////////////////////////////////////////////////////////
//PARA: Date Should In YYYY-MM-DD Format
//RESULT FORMAT:
// '%y Year %m Month %d Day %h Hours %i Minute %s Seconds' =>  1 Year 3 Month 14 Day 11 Hours 49 Minute 36 Seconds
// '%y Year %m Month %d Day'                       =>  1 Year 3 Month 14 Days
// '%m Month %d Day'                                     =>  3 Month 14 Day
// '%d Day %h Hours'                                   =>  14 Day 11 Hours
// '%d Day'                                                 =>  14 Days
// '%h Hours %i Minute %s Seconds'         =>  11 Hours 49 Minute 36 Seconds
// '%i Minute %s Seconds'                           =>  49 Minute 36 Seconds
// '%h Hours                                          =>  11 Hours
// '%a Days                                                =>  468 Days
//////////////////////////////////////////////////////////////////////
function dateDifference($date_1 , $date_2 , $differenceFormat = '%a' )
{
    $datetime1 = date_create($date_1);
    $datetime2 = date_create($date_2);

    $interval = date_diff($datetime1, $datetime2);

    return $interval->format($differenceFormat);

}

仅根据需要设置参数$differenceFormat例如，我希望你的年龄在到岁之间，以月和日为单位

dateDifference（日期（'Y-m-d'），$date，'%Y%m%d'）

或其他格式

dateDifference（日期（'Y-m-d'），$date，'%Y-%m-%d'）