Question: ```
<?php

$dbCon2=mysqli_connect("localhost", "root", "", "dbusers")
       or die(mysql_error()."Connection disconnected");

$sql1 = "SELECT * FROM users";
$sql2 = mysqli_query($dbCon2, $sql1);

    while($row = mysql_fetch_array($sql2))
    {

    echo "<tr>";
        echo "<td>" . $row['UserID'] . "</td>";
        echo "<td>" . $row['Firstname'] . "</td>";
        echo "<td>" . $row['Lastname'] . "</td>";
        echo "<td>" . $row['Gender'] . "</td>";
        echo "<td>" . $row['Email'] . "</td>";
        echo "<td>" . $row['Status'] . "</td>";
        echo "<td>" . $row['Date_joined'] . "</td>";
    echo "</tr>";
    }

?>

```

* I'm still getting the same error again and again. :(

Warning: mysql\_fetch\_array() expects parameter 1 to be resource, object given

Answer: 
That would be because you are mixing `mysql_*` and `mysqli`... two different extensions.

It would appear you're looking for [`mysqli_fetch_array()`](http://www.php.net/manual/en/mysqli-result.fetch-array.php). Also, make sure to use the correct error catching.. you are currently using `mysql_error()` to fetch errors.. I'm not used to using mysqli, but that seems wrong to me.
Change mysql\_fetch\_array to mysqli\_fetch\_array because the returned values from the mysqi\_query is an object and must be passed to a mysqli\_fetch\_array :)