php下统计用户在线时间的一种尝试
下面列出几个比较常用的方法:
首先介绍一下所涉及的数据表结构,四个字段:
复制代码 代码如下:
uid
session_id
login_time
logout_time
1. 客户端定时发送请求到服务器端。实现方法是在用户登录后,将uid,session_id,login_time插入一条记录,然后在客户端js设定一个计时器,比如每10分钟向服务器端发送一个请求,以此来达到更新登出时间的目的,当然这个间隔时间设定的越短,数据可能会越准确,不过相应的系统的负载也会越高,这个可以根据实际情况设定一个合适的值。这种方法广泛应用于webgame上,因为webgame的几乎所有请求都是ajax请求,不用刷新页面,一旦刷新页面,这个计时器就失去了价值,这也是这个方法的局限性。
2. 服务器设定一个定时轮询的脚本。这个方法是在服务器端写一个定时执行的脚本,比如5分钟执行一次,根据数据库中的记录来判断每个会话的session_id是否还存在于服务器上,如果存在就更新logout_time,不存在就跳过。这样也能比较准确的统计在线时间,不过缺点是需要有服务器的控制权,不然无法设定定时脚本,linux系统可以通过crontab实现,windows系统可以通过计划任务来完成。如果你只是买的虚拟主机,那么这个方法也同样不适合你。
3. 在用户每次活动时更新一下登出时间。这样在用户不活动或者退出的时候,登出时间就自然而然的存在于数据库里了,这也是本文着重讨论的方案。下面给出实现方法。
首先,在用户登录成功后,记录下其uid,session_id,并将现在时间作为登陆时间,现在时间+600s作为登出时间,插入数据库。
复制代码 代码如下:
$uid = $_SESSION['uid'] = $info['id'];
$session_id = $_SESSION['session_id'] = session_id();
$login_time = time();
$logout_time = time()+600;
$sql = "INSERT INTO member_login (uid,session_id,login_time,logout_time) values($uid,'$session_id',$login_time,$logout_time)";
mysql_query($sql);
然后在用户每次活动,也就是每点击一个页面时,如果session存在也就是处于登录状态时,更新用户登出时间
复制代码 代码如下:
if($_SESSION['uid']){
$uid = $_SESSION['uid'];
$session_id = $_SESSION['session_id'];
$logout_time = time()+600;
$sql = "UPDATE member_login SET logout_time=$logout_time WHERE uid=$uid AND session_id='$session_id'";
mysql_query($sql);
}
这种方法的优点是相对来说实现起来比较简单,能够适用于大多数的网站,没有额外的服务器需求,而且也可以比较准确的统计用户的在线时间。
缺点也很明显,增加了数据库的更新操作,增加了系统的负载,不过对于中小型网站来说应该不是问题。

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