使用AJAX进行实时搜索

Question

I need to build a live search but i don't know much about ajax and javascript, but i understand html, css and php.Can someone tell me what i'm doing wrong?

<script>
    function liveSearch(str) {
        if (str == "") {
            document.getElementById("txtHint").innerHTML = "";
            return;
        }
        else { 
            if (window.XMLHttpRequest) {
                // code for IE7+, Firefox, Chrome, Opera, Safari
                xmlhttp = new XMLHttpRequest();
            }
            else {
                // code for IE6, IE5
                xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
            }

            xmlhttp.onreadystatechange = function() {
                if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
                    document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
                }
            }

            xmlhttp.open("GET", "getuser.php?q=" + str, true);
            xmlhttp.send();
        }
    }
</script>

<input type="text" onkeypress="liveSearch(this.name)" name="searchWord"/>

<?php
    $dbhost = 'localhost';
    $dbuser = 'root';
    $dbpass = '';
    $dbname = 'ezcart';
    $conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname);

    if ($conn->connect_error) {
        die('Could not connect to the database: ' . $conn->connect_error);
    }

    $word = $_GET['q'];
    $sql= "SELECT prodNam FROM product WHERE prodNam LIKE '$q' ORDER BY prodNam ASC";
    $result = $conn->query($sql);

    if ($result->num_rows > 0) {
        while($row = $result->fetch_assoc()) {
            echo "<div><p>".$row[prodNam]."</p></div>";
        }
    }
    $conn->close();
?>

Answer 1

Change the query

$word = $_GET['q'];

$sql= "SELECT prodNam FROM product WHERE prodNam LIKE '%".$word."%'";

For more information:http://www.w3schools.com/sql/sql_like.asp