Fetch mysql data using ajax when one input is filled

Fetch mysql data using ajax when one input is filled

Respected all users,
      I am facing a little problem in using ajax+jQuery. I have two field in database roll and name . Now i have designed a html form containing two input type field roll and name . I want when i enter the existing roll in the roll field and enter tab or any other key the name field should automatically filled with the related name stored in database.Thanks in advance
My codes are here

index.php
  1. <html>

    <head>
    <script type="text/javascript" src="jquery-1.7.1.min.js"></script>
    <script type="text/javascript">
    var searchTimeout; //Timer to wait a little before fetching the data
    $("#roll").keyup(function() {
        searchKey = this.value;

        clearTimeout(searchTimeout);

        searchTimeout = setTimeout(function() {
            getUsers(searchKey);   
        }, 400); //If the key isn't pressed 400 ms, we fetch the data
    });


    function getUsers(searchKey) {
        $.ajax({
            url: 'getUser.php',
            type: 'POST',
            dataType: 'json',
            data: {value: searchKey},
            success: function(data) {
                if(data.status) {
                    $("#name").val(data.userData.name);
                } else {
                    // Some code to run when nothing is found
                }  
            }
        });        
    }

    </script>


    </head>               
                   
    <table>
        <tr>
            <td>Roll</td>
            <td><input type="text" name="roll" id="roll" /></td>
        </tr>

        <tr>
            <td>Name</td>
            <td><input type="text" name="name" id="name" /></td>
        </tr>
        </table>
        </html>

getUser.php
  1. <?php
        include "db.php";

        $response = Array();

        $response['status'] = false;

        $query = mysql_query("SELECT `name` FROM `tab` WHERE `roll` LIKE '%".$_POST['value']."%' LIMIT 1"); //Or you can use = instead of LIKE if you need a more strickt search

        if(mysql_num_rows($query)) {
            $userData = mysql_fetch_assoc($query);

            $response['userData'] = $userData;
            $response['status'] = true;           
        }

        echo json_encode($response);
        ?>