PHP上次登录脚本

Question

我想知道一个好的,有效的方式来告诉我的用户上次登录的时间.

在用户配置文件中,我想要说明他们上次登录的时间.

例如:

用户1上次登录:2小时前

用户2上次登录:3分钟前

用户3上次登录:2个月前

---

我会将他们的最后登录信息保存在MySQL数据库中,但想知道如何编写脚本.

我刚刚意识到Stackoverflow使用了这个功能,这样可以帮助你理解我想要的东西.

---

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

这就是我将如何更新数据库.

Answer 1

每个页面请求只更新其上一个活动.

<?php

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

要显示他们上次上线的时间,只需从数据库中选择他们的lastactivity字段并显示它

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

relativeTime()我一直在使用的功能在哪里:

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";
}