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'）

对于php版本>=5.3：创建两个日期对象，然后使用date_diff（）函数。它将返回phpDateInterval对象。请参阅文档

$date1=date_create("2007-03-24");
$date2=date_create("2009-06-26");
$diff=date_diff($date1,$date2);
echo $diff->format("%R%a days");

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！

我在PHP5.2中遇到了同样的问题，并用MySQL解决了这个问题。可能并不是你想要的，但这会奏效，并返回天数：

$datediff_q = $dbh->prepare("SELECT DATEDIFF(:date2, :date1)");
$datediff_q->bindValue(':date1', '2007-03-24', PDO::PARAM_STR);
$datediff_q->bindValue(':date2', '2009-06-26', PDO::PARAM_STR);
$datediff = ($datediff_q->execute()) ? $datediff_q->fetchColumn(0) : false;

此处有更多信息http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_datediff

使用date_diff（）尝试这个非常简单的答案，这是经过测试的。

$date1 = date_create("2017-11-27");
$date2 = date_create("2018-12-29");
$diff=date_diff($date1,$date2);
$months = $diff->format("%m months");
$years = $diff->format("%y years");
$days = $diff->format("%d days");

echo $years .' '.$months.' '.$days;

输出为：

1 years 1 months 2 days

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);