使用示例：

echo time_diff_string('2013-05-01 00:22:35', 'now');
echo time_diff_string('2013-05-01 00:22:35', 'now', true);

输出：

4 months ago
4 months, 2 weeks, 3 days, 1 hour, 49 minutes, 15 seconds ago

功能：

function time_diff_string($from, $to, $full = false) {
    $from = new DateTime($from);
    $to = new DateTime($to);
    $diff = $to->diff($from);

    $diff->w = floor($diff->d / 7);
    $diff->d -= $diff->w * 7;

    $string = array(
        'y' => 'year',
        'm' => 'month',
        'w' => 'week',
        'd' => 'day',
        'h' => 'hour',
        'i' => 'minute',
        's' => 'second',
    );
    foreach ($string as $k => &$v) {
        if ($diff->$k) {
            $v = $diff->$k . ' ' . $v . ($diff->$k > 1 ? 's' : '');
        } else {
            unset($string[$k]);
        }
    }

    if (!$full) $string = array_slice($string, 0, 1);
    return $string ? implode(', ', $string) . ' ago' : 'just now';
}

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

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

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;
}

function showTime($time){

    $start      = strtotime($time);
    $end        = strtotime(date("Y-m-d H:i:s"));
    $minutes    = ($end - $start)/60;


    // years 
    if(($minutes / (60*24*365)) > 1){
        $years = floor($minutes/(60*24*365));
        return "From $years year( s ) ago";
    }


    // monthes 
    if(($minutes / (60*24*30)) > 1){
        $monthes = floor($minutes/(60*24*30));
        return "From $monthes monthe( s ) ago";
    }


    // days 
    if(($minutes / (60*24)) > 1){
        $days = floor($minutes/(60*24));
        return "From $days day( s ) ago";
    }

    // hours 
    if(($minutes / 60) > 1){
        $hours = floor($minutes/60);
        return "From $hours hour( s ) ago";
    }

    // minutes 
    if($minutes > 1){
        $minutes = floor($minutes);
        return "From $minutes minute( s ) ago";
    }
}

echo showTime('2022-05-05 21:33:00');

“如果”日期存储在MySQL中，我发现在数据库级别进行差异计算更容易。。。然后根据“天”、“小时”、“分钟”、“秒”输出，分析并显示相应的结果。。。

mysql> select firstName, convert_tz(loginDate, '+00:00', '-04:00') as loginDate, TIMESTAMPDIFF(DAY, loginDate, now()) as 'Day', TIMESTAMPDIFF(HOUR, loginDate, now())+4 as 'Hour', TIMESTAMPDIFF(MINUTE, loginDate, now())+(60*4) as 'Min', TIMESTAMPDIFF(SECOND, loginDate, now())+(60*60*4) as 'Sec' from User_ where userId != '10158' AND userId != '10198' group by emailAddress order by loginDate desc;
 +-----------+---------------------+------+------+------+--------+
 | firstName | loginDate           | Day  | Hour | Min  | Sec    |
 +-----------+---------------------+------+------+------+--------+
 | Peter     | 2014-03-30 18:54:40 |    0 |    4 |  244 |  14644 |
 | Keith     | 2014-03-30 18:54:11 |    0 |    4 |  244 |  14673 |
 | Andres    | 2014-03-28 09:20:10 |    2 |   61 | 3698 | 221914 |
 | Nadeem    | 2014-03-26 09:33:43 |    4 |  109 | 6565 | 393901 |
 +-----------+---------------------+------+------+------+--------+
 4 rows in set (0.00 sec)