"Who is logged in" Not working - php

Question

I'm trying to write a system that shows who is logged into my website.
I've tried so many things, and I feel like I'm onto something with this - but it isn't working. I need help trying to find where I am going wrong.
Here is my display code (I know I shouldn't format with tables, but I am using it for testing):
<?
$loggedInUsers2 = "SELECT * FROM techs WHERE Level='0' AND LastTimeSeen>DATE_SUB(NOW(), INTERVAL 5 MINUTE)";
$loggedInUsers3 = "SELECT * FROM techs WHERE Level>'0' AND LastTimeSeen>DATE_SUB(NOW(), INTERVAL 5 MINUTE)";
?>
<div class="col-sm-4">
<div class="thumbnail">
<center><h4 style="height:35px;">Users Online</h4></center>
<div class="modal-body" style="min-height:498px;">
<table>
<tr><td>
<?
mysqli_query($con, $loggedInUsers2) or die("Error " . mysqli_error($con));
while($row2 = mysqli_fetch_array($loggedInUsers2)) { //if level less than 1
echo $row2['Name']."<br/>";
}
?>
</td><td>
<?
mysqli_query($con, $loggedInUsers3) or die("Error " . mysqli_error($con));
while($row3 = mysqli_fetch_array($loggedInUsers3)) { //if level more than 1
echo $row3['Name']."<br/>";
}
?>
</td></tr>
</table>
</div>
</div>
</div>
Here is my saving to the database:
$userId = $_SESSION['UserId'];
$loggedInUsers1 = "UPDATE techs SET LastTimeSeen=NOW() WHERE UniqueID='$userId'";
mysqli_query($con, $loggedInUsers1) or die("Error " . mysqli_error($con));
This outputs to my database (In the LastTimeSeen field) to something like 2015-12-21 08:35:43 (Updates every few seconds via jquery reloading the footer of the page)
Basically, there is just no output in the tables from the first page.
EDIT:
It was suggested to use the login button to set a user active, and the logout button to set them inactive - here is my response to that:
"The problem with that is people won't use the logout button. They will just close the browser. I want this to keep track of only users that are online. The footer updates the time in the database, every couple of seconds, and then the table is listed on a part of the page that reloads every few seconds aswell, so they are both always up to date. It should only list people that have been on a page in the past 5 minutes."

Answer 1

When you use mysqli_query it returns a result object. You need to pass this object into mysqli_fetch_array not the query string.
<?
$loggedInUsers2 = "SELECT * FROM techs WHERE Level='0' AND LastTimeSeen>DATE_SUB(NOW(), INTERVAL 5 MINUTE)";
$loggedInUsers3 = "SELECT * FROM techs WHERE Level>'0' AND LastTimeSeen>DATE_SUB(NOW(), INTERVAL 5 MINUTE)";
?>
<div class="col-sm-4">
<div class="thumbnail">
<center><h4 style="height:35px;">Users Online</h4></center>
<div class="modal-body" style="min-height:498px;">
<table>
<tr><td>
<?
$results1 = mysqli_query($con, $loggedInUsers2) or die("Error " . mysqli_error($con));
while($row2 = mysqli_fetch_array($results1)) { //if level less than 1
echo $row2['Name']."<br/>";
}
?>
</td><td>
<?
$results2 = mysqli_query($con, $loggedInUsers3) or die("Error " . mysqli_error($con));
while($row3 = mysqli_fetch_array($results2)) { //if level more than 1
echo $row3['Name']."<br/>";
}
?>
</td></tr>
</table>
</div>
</div>
</div>

Answer 2

I think because last seen = now is only equal to NOW when the query runs no? meaning that its set to now for only the split second? i could be wrong but throwing ideas out there. to be honest I think there is an easier way to do this aswell!
When you log a user in also run a query to update a field.. so on log in
//Code to log in
if password and email comp is found in database then log user in
//Code to set user active
UPDATE user WHERE username& password match, SET active = TRUE or 1 or something of the sort.
Then on logout set active = to false or 0, I hope you see the logic im trying to explain here, I would write the code but been too deep in laravel and not touched vanilla db management for a while now.

Answer 3

I think this task shouldn't be done with MySQL at all. Because you're going to face a huge overload and database deadlocks when there will be thousands of visitors opening a new page every second.
You would be better off by using a RAM-based database, like Redis (http://redis.io).
In Redis, you can set automatically expiring entries (let's say, for 5 minutes) for currently active visitors. If that entry is not updated for 5 minutes, it would be automatically deleted. I would suggest to use Redis command SETEX for that.
You wouldn't need to worry about people who are not logging out and you would achieve exactly what you want in an efficient way.

Answer 4

So as spotted earlier you will want to change the following on both lines of code.
mysqli_query($con, $loggedInUsers2) or die("Error " . mysqli_error($con));
while($row2 = mysqli_fetch_array($loggedInUsers2)) { //if level less than 1
echo $row2['Name']."<br/>";
}
TO
$query_1 = mysqli_query($con, $loggedInUsers2) or die("Error " . mysqli_error($con));
while($row2 = mysqli_fetch_array($query_1)) { //if level less than 1
echo $row2['Name']."<br/>";
}
The problem is that the parameters passed into the mysqli_fetch_array is not the query object, Meaning the loop was just going through a MySQL statement not the object.