Trying to get my live search to work ‘properly’. When I enter an ‘a’

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Asked: June 18, 20262026-06-18T02:04:14+00:00 2026-06-18T02:04:14+00:00

Trying to get my live search to work ‘properly’. When I enter an ‘a’

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Trying to get my live search to work ‘properly’. When I enter an ‘a’ I get ‘aussie’ and ‘australian’ (not sure where these are coming from as they are not in my db) in a area below the box (like a drop down). I can select one of them and it populates the text box. The code suggested to me uses mysqli which I am not familiar with. When I add a typical SELECT statement I am use to I see the correct data (more or less) but it is not selectable. The reason I say more or less is an ‘a’ gives me ‘aussie’ and ‘australian’ like before (in a drop down). Adding a ‘n’ (i.e. searching for ‘an’) gives me list of all names in my database that contain the string in a DIV but not selectable.

index.html

<!DOCTYPE html>
<html lang="en">
<head>
    <script type="text/javascript" src="jquery.js"></script>
    <script type="text/javascript"> 
        $(document).ready(function() {
            $("#faq_search_input").keyup(function() {
                var faq_search_input = $(this).val();
                var dataString = 'keyword='+ faq_search_input;

                if(faq_search_input.length>1) {
                    $.ajax({
                        type: "GET",
                        url: "ajax-search.php",
                        data: dataString,
                        success: function(server_response) {
                            document.getElementById("searchresultdata").style.display = "block";
                            $('#searchresultdata').html(server_response).show();
                        }
                    });
                } return false;
            });
        });
    </script>


</head>
<body>
    <div class="searchholder">
       <input  name="query" class="quicksearch" type="text" id="faq_search_input" />
       <div id="searchresultdata" class="searchresults" style="display:none;"> </div>
    </div>
</body>
</html>

ajax-search.php

<?php
    //you must define your database settings
    define("DB_HOST", "localhost");
    define("DB_USERNAME", "root");
    define("DB_PASSWORD", "");
    define("DB_NAME", "mydb");
    if(isset($_GET['keyword'])) {
        $search = new mysqli(DB_HOST, DB_USERNAME, DB_PASSWORD, DB_NAME);
        if ($search->connect_errno) {
            echo "Failed to connect to MySQL: (" . $search->connect_errno . ") " .     $search->connect_error;
            $search->close();
        }
        $keyword =  trim($_GET['keyword']) ;

        //original statement
        //$query ="SELECT COLUMN_NAME FROM ".DB_NAME.".INFORMATION_SCHEMA.COLUMNS WHERE COLUMN_NAME LIKE '%".$keyword."%'";

        //my edited statement
        //$query ="SELECT name FROM ".DB_NAME.".INFORMATION_SCHEMA.COLUMNS WHERE name LIKE '%".$keyword."%'";

        //basic sql statement
        $query ="SELECT name FROM users WHERE name LIKE '%".$keyword."%'";

        $values = $search->query($query);
        if( $search->error ) exit( $search->error ); 

        if($values->num_rows != 0) {
            while($row = $values->fetch_assoc()) { 
                echo $row['name']."<br>";
            } 
        }
        else {
            echo 'No Results for :"'.$_GET['keyword'].'"';
        }
    }
?>

database contains a table of user data called users with a column called name.
Any thoughts?