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

前段时间，我编写了一个format_date函数，因为它提供了许多关于日期的选项：

function format_date($date, $type, $seperator="-")
{
    if($date)
    {
        $day = date("j", strtotime($date));
        $month = date("n", strtotime($date));
        $year = date("Y", strtotime($date));
        $hour = date("H", strtotime($date));
        $min = date("i", strtotime($date));
        $sec = date("s", strtotime($date));

        switch($type)
        {
            case 0:  $date = date("Y".$seperator."m".$seperator."d",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 1:  $date = date("D, F j, Y",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 2:  $date = date("d".$seperator."m".$seperator."Y",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 3:  $date = date("d".$seperator."M".$seperator."Y",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 4:  $date = date("d".$seperator."M".$seperator."Y h:i A",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 5:  $date = date("m".$seperator."d".$seperator."Y",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 6:  $date = date("M",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 7:  $date = date("Y",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 8:  $date = date("j",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 9:  $date = date("n",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 10: 
                     $diff = abs(strtotime($date) - strtotime(date("Y-m-d h:i:s"))); 
                     $years = floor($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));
                     $date = $years . " years, " . $months . " months, " . $days . "days";
        }
    }
    return($date);
}    

您可以始终使用以下函数，以年和月为单位返回年龄（即1年4个月）

function getAge($dob, $age_at_date)
{  
    $d1 = new DateTime($dob);
    $d2 = new DateTime($age_at_date);
    $age = $d2->diff($d1);
    $years = $age->y;
    $months = $age->m;

    return $years.'.'.months;
}

或者如果希望在当前日期计算年龄，可以使用

function getAge($dob)
{  
    $d1 = new DateTime($dob);
    $d2 = new DateTime(date());
    $age = $d2->diff($d1);
    $years = $age->y;
    $months = $age->m;

    return $years.'.'.months;
}

您可以使用

getdate()

函数，该函数返回包含所提供日期/时间的所有元素的数组：

$diff = abs($endDate - $startDate);
$my_t=getdate($diff);
print("$my_t[year] years, $my_t[month] months and $my_t[mday] days");

如果开始和结束日期为字符串格式，则使用

$startDate = strtotime($startDateStr);
$endDate = strtotime($endDateStr);

在上述代码之前

$date1 = date_create('2007-03-24');
$date2 = date_create('2009-06-26');
$interval = date_diff($date1, $date2);
echo "difference : " . $interval->y . " years, " . $interval->m." months, ".$interval->d." 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);