PHP Count Logged-In Users

Question

I\'m having trouble figuring out how I can count the number of logged-in users in my application.

What I have: When a user logs in, they get a sessi

Answer 1

Just to offer another solution:

if ($user->isLoggedIn()) {
  touch("/writable/path/loggedInUsers/" . $user->id);
}

If you don't need to query this data, a local file touch is far faster than a DB write. To get logged in users, scan the directory for filemtimes under N seconds old.

Answer 2

Rather than a IsLoggedIn column, you should add a LastTimeSeen column. Any time a person visits a page you update the column:

UPDATE members SET LastTimeSeen = NOW() WHERE id = $the_user_id

Then to get how many people are on the site at any given moment you use the query:

SELECT COUNT(*) FROM members WHERE LastTimeSeen > DATE_SUB(NOW(), INTERVAL 5 MINUTE)

That shows how many people have viewed a page in the past 5 minutes, which is the best you're gonna get without a much more complicated solution.

Answer 3

Because of the way our site is constructed, it was necessary to use the ajax approach. I'm using jQuery so it's relatively painless.

These lines went into the $(document).ready function.

fnShowImOnline();
setInterval('fnShowImOnline', 120000);

This is the javascript function...

function fnShowImOnline() {
    $.get('ajax/im_online.php');
}

And here is the PHP

<?php
    session_start();
    if ((isset($_SESSION['user']))&&($_SESSION['authorized']=='authorized')) {
        include('../includes/db.php');
        db_connect();
        mysql_query("UPDATE members SET last_checked_in = NOW() WHERE user_id = {$_SESSION['user']['user_id']}");
    }

?>

The count is straight PHP/mySQL.

//  Members online.
$online_sql = "SELECT COUNT(*) FROM members where last_checked_in > DATE_SUB(NOW(), INTERVAL 5 MINUTE)";
$online_RS = mysql_query($online_sql);
$online_row = mysql_fetch_row($online_RS);
$online = $online_row[0];

For those times I need to update the numbers dynamically, this bit of ajax does the trick.

$.ajax({
    url: 'ajax/members_online.php',
    dataType: 'json',
    success: function(response) {
        if (!isNaN(response.total)) {
            $('#OnlineTotal').html(response.total + " Total ");
            $('#OnlineOnline').html(response.online +  " Online Now");
        }
    }
})

using this for the PHP/mySQL

//  Members online.
$online_sql = "SELECT COUNT(*) FROM members WHERE last_checked_in > DATE_SUB(NOW(), INTERVAL 5 MINUTE)";
$online_RS = mysql_query($online_sql);
$online_row = mysql_fetch_row($online_RS);
$online = $online_row[0];
//  Members total.
$total_sql = "SELECT COUNT(*) FROM members";
$total_RS = mysql_query($total_sql);
$total_row = mysql_fetch_row($total_RS);
$total = $total_row[0];
$response = json_encode(array('total'=>$total,'online'=>$online));
echo($response);

This is working well for us.