Php 如何显示上次登录时间（年、月、周、日、时间、秒）。。我试过了，但输出错误

$lastLoginTime=日期（“Y-m-d H:i:s”）//登录期间将存储在数据库中的最后一次登录

假设上次登录是：2012-05-11 07:57:46
当前时间为：$Current_Time=date（“Y-m-d H:i:s”）

输出错误。。。。！ ：-（

谷歌（试一下）给了我这个：

<?php
function duration($seconds, $max_periods) {
    $periods = array("year" => 31536000, "month" => 2419200, "week" => 604800, "day" => 86400, "hour" => 3600, "minute" => 60, "second" => 1);
    $i = 1;
    foreach ( $periods as $period => $period_seconds )
    {
        $period_duration = floor($seconds / $period_seconds);
        $seconds = $seconds % $period_seconds;
        if ( $period_duration == 0 ) continue;
        $duration[] = "{$period_duration} {$period}" . ($period_duration > 1 ? 's' : '');
        $i++;
        if ( $i >  $max_periods ) break;
    }
    return implode(' ', $duration);
}
$diff_seconds  = strtotime($current_time) - strtotime($lastLoginTime);

print(duration($diff_seconds, 6));
?>

---

好的..谢谢..我认为$diff_seconds=strotime（$current_time）-strotime（$lastLoginTime）；上次登录是：2012-05-11 10:16:51当前时间是：2012-05-11 10:35:51$diff_seconds=strotime（$current_time）-strotime（$lastLoginTime）；如果您没有显示变量$current_time和$lastlogentime的内容，那么$diff_seconds-13367241as的ans是否正确。请转储它们并告诉：-）

<?php
function duration($seconds, $max_periods) {
    $periods = array("year" => 31536000, "month" => 2419200, "week" => 604800, "day" => 86400, "hour" => 3600, "minute" => 60, "second" => 1);
    $i = 1;
    foreach ( $periods as $period => $period_seconds )
    {
        $period_duration = floor($seconds / $period_seconds);
        $seconds = $seconds % $period_seconds;
        if ( $period_duration == 0 ) continue;
        $duration[] = "{$period_duration} {$period}" . ($period_duration > 1 ? 's' : '');
        $i++;
        if ( $i >  $max_periods ) break;
    }
    return implode(' ', $duration);
}
$diff_seconds  = strtotime($current_time) - strtotime($lastLoginTime);

print(duration($diff_seconds, 6));
?>