Users online script, where is the problem?

I have a users online script that tells me the users that are logged in. There is only one problem. Well, its not an error. But take a look:

Just checking to see if the user is online

function chklogin($postuser) {
$time = time(); 
$previous = "180"; 
$timeout = $time-$previous; 
$online = mysql_query("SELECT * FROM online WHERE timeout > '$timeout'"); 
$row_online = mysql_fetch_array($online); 
if ($row_online['username'] == $postuser) { 
return '<img src="/images/icons/user_comment.png" alt="online" height="16" width="16" />';
} else {
return '<img src="/images/icons/user_delete.png" alt="offline" height="16" width="16" />'; 
}
}

PHP:

This bit, shows all of the users online, not just one.

$time = time(); 
$previous = "180"; 
$timeout = $time-$previous;

if (isset($_SESSION['username'])) { 
$username = $_SESSION['username']; 
$verify = mysql_query("SELECT * FROM online WHERE username=\"$username\" AND timeout > \"$timeout\""); 

$row_verify = mysql_fetch_assoc($verify); 
if (!isset($row_verify['username'])) { 
$insert = mysql_query("INSERT INTO online (username, timeout) VALUES (\"$username\", \"$time\")"); 
}
}

$online = mysql_query("SELECT * FROM online WHERE timeout > \"$timeout\""); 
$row_online = mysql_fetch_assoc($online); 
if (isset($row_online['username'])) { 
echo 'Logged in users: ';
do {
echo $row_online['username'].", "; 
} while($row_online = mysql_fetch_assoc($online));
} else {
echo "There are no logged in users."; 
}

PHP:

The second one works. On the first one, I am using it on my forum. For each post, it has a chklogin($username); The problem is that when there is 2 or more users online, it only says that one is online. It says all the others are offline.

::Notice::! When I say "it only says that one is online" I means the first bit on code. That has the function chklogin();

 

Well from what I can tell you should be looping through all the MySQL results, not just the first one. It seems you just get the first result, check to see if it matches the current user and if it does then you say they are online.

Basically just add a while loop to the check:

$result= mysql_query("SELECT * FROM online WHERE timeout > '$timeout'");
while ($row = mysql_fetch_assoc($result))
{
    if ($row['username'] == $postuser) { 
        return '<img src="/images/icons/user_comment.png" alt="online" height="16" width="16" />';
    } else {
        return '<img src="/images/icons/user_delete.png" alt="offline" height="16" width="16" />'; 
    }
}

PHP:

 

Wouldn't it be a better idea to query the number of rows as opposed to retreiving and looping over the entire table?

 

I just want to grab the if(user is online) data for 1 user

 

No idea what you're talking about. He needs to check the username for each result in which timeout has not yet expired so it's only logical to loop through each result.

 

I'd do it something like this (unless I've got the wrong end of the stick so-to-speak):

<?php
mysql_query('SELECT `id` FROM `online` WHERE `username` = "' . $postuser . '" AND `timeout` > ' . $timeout); // replace the id field name appropriately, there is no need to retreive all fields, one will do
$iRows  = mysql_num_rows($rQuery);
mysql_free_result($rQuery);

if ($iRows == 1)
{
	// online
}
else
{
    // offline
}
?>

PHP:

 

Ok, you're not understanding my point. Did you try my code? I think if you do you'll see what I'm saying.

Think about what's happening. You're querying the database to retrieve ALL timeouts that are greater than the current timeout. Assuming there's not one row in your database, multiple rows can be returned. If you want to just find one user then just query like this:

$online = mysql_query("SELECT * FROM online WHERE timeout > '$timeout' AND username = '$postuser'");

PHP:

 

Alright AliasXNeo, I tried that one line, and it worked. Thanks to you two for helping

 

If I understand and you only need the data from one user, the current method is overkill.

My method is more efficient and quicker.

 

Yes your method would be cleaner and I would have suggested it if I understood the first time.

The best way to do this:

if (mysql_num_rows(mysql_query("SELECT `username` FROM `online` WHERE `timeout` > '{$timeout}' AND `username` = '{$postuser}'"))
{
    // User is active //
} else {
    // User is not active //
}

PHP:

Make sure you use mysql_real_escape_string() on $postuser for security reasons.

 

Hmm, I dont think so, I protected that in the registration.

Okay, So this should work out for me