PHP Last登录脚本 [英] PHP Last log in script

问题描述

我想知道一种有效且有效的方法，以判断用户上次登录的时间.

I wanted to know a good and efficient way to be able to tell how long ago my users last logged in.

在用户个人资料上，我想说一下他们的上一次登录是多久之前的.

On the users profile I want it for say how long ago their last log in was.

例如:

用户1 上次登录:2小时前

User 1 Last login: 2 hours ago

用户2 上次登录时间:3分钟前

User 2 Last login: 3 minutes ago

用户3 上次登录时间:2个月前

User 3 Last login: 2 months ago

我会将他们的上次登录信息保存在MySQL数据库中，但想知道如何执行脚本.

I will keep their last login information in a MySQL database but want to know how to do the script.

我刚刚意识到Stackoverflow使用了此功能，因此可以帮助您了解我想要的内容.

I just realized that Stackoverflow uses this feature, so that can help you understand what I want.

mysql_query("UPDATE users SET lastactivity = ".time()." WHERE id = ".$userID);

这就是我更新数据库的方式.

This is how I will update the DB.

推荐答案

每个页面请求仅更新其上一个活动.

Each page request just update their last activity.

<?php

mysql_query("UPDATE users SET lastactivity = ".time()." WHERE id = ".$userID);

要显示他们上次上网的时间，只需从数据库中选择其lastactivity字段并显示

to show when they were last online, just select their lastactivity field from database and show it

<?php
$activity = mysql_result(mysql_query("SELECT lastactivity FROM users WHERE id = ".$userID), 0);
echo "Last activity: ".relativeTime($active);

其中relativeTime()是我一直在使用的功能:

where relativeTime() is function I've been using:

function relativeTime($time, $short = false){
    $SECOND = 1;
    $MINUTE = 60 * $SECOND;
    $HOUR = 60 * $MINUTE;
    $DAY = 24 * $HOUR;
    $MONTH = 30 * $DAY;
    $before = time() - $time;

    if ($before < 0)
    {
        return "not yet";
    }

    if ($short){
        if ($before < 1 * $MINUTE)
        {
            return ($before <5) ? "just now" : $before . " ago";
        }

        if ($before < 2 * $MINUTE)
        {
            return "1m ago";
        }

        if ($before < 45 * $MINUTE)
        {
            return floor($before / 60) . "m ago";
        }

        if ($before < 90 * $MINUTE)
        {
            return "1h ago";
        }

        if ($before < 24 * $HOUR)
        {

            return floor($before / 60 / 60). "h ago";
        }

        if ($before < 48 * $HOUR)
        {
            return "1d ago";
        }

        if ($before < 30 * $DAY)
        {
            return floor($before / 60 / 60 / 24) . "d ago";
        }


        if ($before < 12 * $MONTH)
        {
            $months = floor($before / 60 / 60 / 24 / 30);
            return $months <= 1 ? "1mo ago" : $months . "mo ago";
        }
        else
        {
            $years = floor  ($before / 60 / 60 / 24 / 30 / 12);
            return $years <= 1 ? "1y ago" : $years."y ago";
        }
    }

    if ($before < 1 * $MINUTE)
    {
        return ($before <= 1) ? "just now" : $before . " seconds ago";
    }

    if ($before < 2 * $MINUTE)
    {
        return "a minute ago";
    }

    if ($before < 45 * $MINUTE)
    {
        return floor($before / 60) . " minutes ago";
    }

    if ($before < 90 * $MINUTE)
    {
        return "an hour ago";
    }

    if ($before < 24 * $HOUR)
    {

        return (floor($before / 60 / 60) == 1 ? 'about an hour' : floor($before / 60 / 60).' hours'). " ago";
    }

    if ($before < 48 * $HOUR)
    {
        return "yesterday";
    }

    if ($before < 30 * $DAY)
    {
        return floor($before / 60 / 60 / 24) . " days ago";
    }

    if ($before < 12 * $MONTH)
    {

        $months = floor($before / 60 / 60 / 24 / 30);
        return $months <= 1 ? "one month ago" : $months . " months ago";
    }
    else
    {
        $years = floor  ($before / 60 / 60 / 24 / 30 / 12);
        return $years <= 1 ? "one year ago" : $years." years ago";
    }

    return "$time";
}

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