检测，如果两个浏览器窗口打开同一个网站

我试图来检测是否用户有我的网站在一个标签页中打开，然后在另一个选项卡中打开它。这应该然后显示一个警告，新开卡的用户。

I'm trying to detect if user has my website open in one tab, then opens it in another tab. This should then show a warning to the user on the newly opened tab.

现在，我发送一个保持活动Ajax调用每秒回服务器。这则记录在一个唯一的ID数据库中的最后一个活动的时间。当一个新的页面被加载它会检查，如果用户（通过用户ID识别）是活跃在最后2秒之内。

Right now I'm sending a "keep alive" ajax call every second back to the server. This then logs the "last active" time in the database with a unique ID. When a new page is loaded it checks if that user (identified by a userID) was active within the last 2 seconds.

这感觉非常低效的。它也不能很好地执行，当用户刷新页面。 （这需要不到2秒）。我做了2秒钟的检查长于存活呼吁允许较慢的连接。

This feels very inefficient. It also doesn't perform well when the user refreshes the page. (which takes less than 2 seconds). I've made the 2 second check longer than the keepalive call to allow for slower connections.

Q）我要对这个正确的方式还是有这样做的更简单的方法？我听到的Gmail和Facebook可能会做同样的事情，但我无法找到该网站的功能中，他们做到这一点。
中国 实际code使用的内部功能很多所以这里的总体思路：

Q) Am I going about this the right way or is there an easier way of doing this? I heard Gmail and Facebook might do something similar but I can't find the site function in which they do it.


Actual code uses lots of in-house functions so here's the general idea:

伪code：

Page load;
[index] PHP: Save tabsessionid, userid, datecreated, lastactive, phpsessionid into database;
[index] JS: Every second, send tabsessionid to keepalive.php
[keepalive] PHP: Check if another session exists in the last 2 seconds
[keepalive] PHP:        If exists -> Return "-1"; else -> return "1";
[keepalive] PHP: Update tabsessionid's last active time.

我已经尝试了一些解决方案的here但警告新的标签，而不是旧的人似乎是在压低延迟/时间方面的棘手的部分。

I've tried some of the solutions here but warning the newer tab, not the older one seems to be the tricky part in terms of keeping down latency/time.