PHP - Appending To Variable

<script type="text/javascript">
jQuery(function($) {

  var rowCtr = 0;
  $('.addRow').click(function(){
		$('#tableID > tbody:last').append('<tr name="rowEq['+rowCtr+']" id="rowEq['+rowCtr+']"><td>'+
				'<select id="stype'+rowCtr+'" name="rowData['+rowCtr+'][equipment_type]">'+
					'<option value="0"> -- </option>'+
					'<option value="1"> Type 1 </option>'+
					'<option value="2"> Type 2 </option>'+
					'<option value="3"> Other </option>'+
				'</select>'+
				'<input id="other_type'+rowCtr+'" type="text" name="rowData['+rowCtr+'][other_type]">'+
				);
				
				$('#stype'+rowCtr).change(function() {
			
				if(jQuery('#stype'+rowCtr).val() == '3')
				{
					jQuery('#other_type'+rowCtr).show();
				}else{
					jQuery('#other_type'+rowCtr).hide();
				}
				});
				
				if(jQuery( '#stype'+rowCtr ).val() != '3')
				{
					jQuery('#other_type'+rowCtr).hide();
				}
				rowCtr++;
   });

});

				

</script>

<input type='button' class='addRow' value="Add Row">
<form>
    <table id="tableId">
       <thead>
           <tr>
               <th> Type </th>
           </tr>
       </thead>
        <tbody>
            <tr>
                <td>
                  dummy text..  
                </td>
            </tr>
        </tbody>
    </table>
</form>

My login script stores the user's login name as $_SESSION[ 'name'] on login. For some unapparent reason, i'm getting errors stating that $user  and $priv are  undefined variables, though I've attempted to define $user as being equal to $_SESSION['name'], using $user to look up the the user's privilege level (stored as the su column ) in the SQL table, and then where the result of the sql query is $priv which is then evaluated in an if statement. I can't seem to figure out why this might not be working.

The code I'm using:

<?php
session_start();
function verify() {
	//verify that the user is logged in via the login page. Session_start has already been called. 
	if (!isset($_SESSION['loggedin'])) {
	header('Location: /index.html'); 
	exit;
	}
	//if user is logged in, we then lookup necessary privleges. $_SESSION['name'] was written with the login name upon login. Privleges   // are written in db as a single-digit integer of of 0 for users, 1 for administrators, and 2 for special users. 
        $user === $_SESSION['name'];
    //Connect to Databse
	$link = mysqli_connect("127.0.0.1", "database user", "password", "database");
	if (!$link) {
  	  echo "Error: Unable to connect to MySQL." . PHP_EOL;
    	echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
    	echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
    	exit;
        }
        //SQL Statement to lookup privlege information. 
       if ($result = mysqli_query($link, "SELECT su FROM accounts WHERE username = $user", MYSQLI_STORE_RESULT)) {
         //LOOP TO CYCLE THROUGH SQL RESULTS AND STORE Privlege information as vairable $priv.
            while ($row = $result->fetch_assoc()) {
           $priv === $row["su"];
            }
        }
        // close SQL connection.
        mysqli_close($link);

 // Verify privleges and take action. Only a privlege of "1" is allowed to view this page. A privlege of "2" indicates special 
//accounts used in other scripts that have certain indermediate additional functions, but are not trusted administrators. 

        if ($priv !== 1) {
        echo $_SESSION['name'];
        echo "you have privlege level of $priv";
        echo "<br>";
        echo 'Your account does not have the privleges necessary to view this page';
        exit;
        }
        
}
verify();
?>

I have a function that get's a quick single item from a query:

function gimme($sql) { 
global $mysqli; 
global $mytable; 
global $sid; 
$query = "SELECT ".$sql." FROM ".$mytable." WHERE sid = ".$sid; $result = $mysqli->query($query); 
$value = $result->fetch_array(MYSQLI_NUM); 
$$sql = is_array($value) ? $value[0] : ""; 
return $$sql; // this is what I've tried so far 
$result->close(); 
}

It works great as:

echo(gimme("name"));

Then I realized that I could use that as a variable ('$name' in this case) elsewhere. However, I can't figure out how get that new, variable variable 'outside' of the function.  As such,

echo($name);

isn't working outside the function.  Is there a way to return a variable variable?  In other words, is there a way to make a function that creates a variable variable that will available outside of the function?

Thanks