Ajax Not Retrieving Elements from Database

Question

I have a dropdown menu that I populate with items from a database. I want to populate text fields with other attributes from the same table I get my dropdown options from. The problem I'm having is retrieving the other attributes, and placing them in the text field. I have written console logs to help me see what my code is doing. At this point, every console log for every function gets returned as if the code is working as it should. The only thing that's not happening is when my code goes to the ajax processing file, it is not getting the data, so when it goes back to the main file, there is no attributes to place in the text fields.

    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>

    <?php                        
    # Perform database query
    $query = "SELECT * FROM student";
    $result = $conn->query($query) or die('Query 1 failed: ' . mysql_error());

    ?>

    <label for="studentSelect">Student Name: </label>
    <select id="studentSelect">
    <option value="0">Please select</option>
    <?php
    while ($row = $result->fetch_assoc())
    {
        echo '<option value="' . $row['studentID'] . '" > "' . $row['studentFirstName'] . '"  "' . $row['studentLastName'] . '"</option>';
    }
    ?>
    </select>
    <div>
        <label for="element_5_1">First Name</label>
        <input id="element_5_1" name="element_5_1" class="element text large" type="text">
    </div>

    <div>
        <span class="floatLeft">
        <label for="element_5_3">Last Name</label>
        <input id="element_5_3" name="element_5_3" class="element text medium" style="width:14em" type="text">
        </span>

        <span style="float:left">
        <label for="element_5_4">Major</label>
        <input id="element_5_4" name="element_5_4" class="element text medium" style="width:4em" type="text">
        </select>
        </span>
        <span style="float:left">
        <label for="element_5_5">Credits Earned</label>   
        <input id="element_5_5" name="element_5_5" class="element text medium" style="width:6em" type="text">
        </span>
    </div>



    <script type="text/javascript">
    function makeAjaxRequest(studentFirstName)
    {
        console.log("making request");
        $.ajax({        

        type: "POST",
        data: { studentFirstName: studentFirstName },
        dataType: "json", 
        url: "process_ajax.php",
        success: function(json) 
        {
            insertResults(json);
            console.log("successful post");
        },
        failure: function (errMsg) 
        {
            alert(errMsg);
        }
        });
        }                               

        $("#studentSelect").on("change", function()
        {       
            var id = $(this).val();
            if (id === "0")
            {
                clearForm();
                console.log("cleared form");
            }
            else 
            {
                makeAjaxRequest(id);
            }
        });                             

        function insertResults(json)
        {                                   

            $("#element_5_1").val(json["studentFirstName"]);

            $("#element_5_3").val(json["studentLastName"]);

            $("#element_5_4").val(json["major"]);

            $("#element_5_5").val(json["creditsEarned"]);
        }

        function clearForm()
        {
            console.log("in clearForm");
            $("#element_5_1, #element_5_3, #element_5_4, #element_5_5").val("");
        }

    </script>

I then have a separate ajax processing file

<?php
$host = "********.mysql.database.azure.com";
$username = "************";
$password = "*******";
$db_name = "**********";

//Establishes the connection
$conn = mysqli_init();
mysqli_real_connect($conn, $host, $username, $password, $db_name, 3306);
if (mysqli_connect_errno($conn)) {
die('Failed to connect to MySQL: '.mysqli_connect_error());

$studentName = $_POST['studentFirstName'];

$query = "SELECT * FROM student";
$result = mysql_query($query) or die('Query 2 failed: ' . mysql_error());
while ($row = mysql_fetch_assoc($result)) {
    if ($studentName == $row['studentFirstName']){
        echo json_encode($row);
    }
}
?> 

Any help at all is much appreciated!