了解有多少用户在PHP在线？ [英] Find out how many users are online in PHP?

问题描述

每次访问我的网站都会更新用户的个人点击计数器，并根据Cookie中存储的IP地址和ID更新 time（）的列。因此，当输出数据时，下面的代码使用更少的数据库调用是更有效的方法，因为它本质上是一个自身的副本：

Every visit to my website updates a user's individual hit counter and updates a column for time() based on their ip address and id stored in a cookie. So when coming to output the data, what's a more efficient way of my following code with less database calls, as it's essentially a copy of itself:

<?
$last1Min = time()-60;
$last5Mins = time()-300;
$last1Hr = time()-6000;
$last1Dy = time()-144000;
$last1Wk = time()-1008000;
$last1Mnth = time()-30240000;

//last1Min
$sql = "SELECT COUNT(*) FROM usersonline WHERE lastOnline > $last1Min";
while($rows = mysql_fetch_array(mysql_query($sql))) {
    echo "Users online in the last minute: " . $rows['COUNT(*)'] . "<br />\n";
}

//last5Mins
$sql = "SELECT COUNT(*) FROM usersonline WHERE lastOnline > $last5Mins";
while($rows = mysql_fetch_array(mysql_query($sql))) {
    echo "Users online in the last 5 minutes: " . $rows['COUNT(*)'] . "<br />\n";
}

//last1Hr
$sql = "SELECT COUNT(*) FROM usersonline WHERE lastOnline > $last1Hr";
while($rows = mysql_fetch_array(mysql_query($sql))) {
    echo "Users online in the last hour: " . $rows['COUNT(*)'] . "<br />\n";
}

//last1Dy
$sql = "SELECT COUNT(*) FROM usersonline WHERE lastOnline > $last1Dy";
while($rows = mysql_fetch_array(mysql_query($sql))) {
    echo "Users online in the last day: " . $rows['COUNT(*)'] . "<br />\n";
}

//last1Wk
$sql = "SELECT COUNT(*) FROM usersonline WHERE lastOnline > $last1Wk";
while($rows = mysql_fetch_array(mysql_query($sql))) {
    echo "Users online in the last week: " . $rows['COUNT(*)'] . "<br />\n";
}

//last1Mnth
$sql = "SELECT COUNT(*) FROM usersonline WHERE lastOnline > $last1Mnth";
while($rows = mysql_fetch_array(mysql_query($sql))) {
    echo "Users online in the last month: " . $rows['COUNT(*)'] . "<br /><br />\n";
}

如果有更有效的方式呈现此数据，将其扩展为不仅显示每个指标的用户数在我的整个网站在线，而且记录和输出我网站上每个网页的数据。

If there is a more efficient way of presenting this data, I'm wanting to extend it to show not only how many users for each of these metrics is online on my entire site, but record and output the data for every page on my site.

推荐答案

SELECT 
  SUM(lastOnline <= 60) AS one_minute,
  SUM(lastOnline <= 300) AS five_minutes,
  ...
  SUM(lastOnline <= 30240000) AS one_month
FROM usersonline

b $ b

使用此方法，您可以通过单个表扫描在单个查询中获得所需的所有内容;它没有得到比这更高效的。正如其他人提到的，你应该缓存结果，因为它相对昂贵（即使在这种优化的形式）。在每次网页加载时，计算这一点没有意义，特别是如果你每秒看到多次点击（这很有可能，如果你打到digg的首页）。

Using this method, you can get everything you need in a single query with a single table scan; it doesn't get much more efficient than that. As others have mentioned, you should cache the result, as it's relatively expensive (even in this optimized form). There's no point in calculating this on every page load, especially if you're seeing multiple hits per second (which is extremely likely if you, say, hit the front page of digg)

lastOnline <= 60求值为1，对于条件为false的行，值为0; SUM（）将这些1和0求和，得到条件为真的行数。

lastOnline <= 60 evaluates to 1 for rows where the condition is true, and 0 for rows where the condition is false; SUM() sums these 1s and zeros, giving you a count of the number of rows for which the condition is true.

从mysql中的用户注释学习这个技术docs几年前;有其他地方的类似示例

Learned this technique from a user comment in the mysql docs a few years ago; there are similar examples elsewhere

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