php - 使用 AJAX 完成自动预测 - 不工作

Question

AJAX JS 功能存在问题，该功能将从 PHP 数据库中获取信息并在用户完成之前预测搜索词。

/js/primary.js：

$(document).ready(function() {
                $('.autosuggest').keyup(function() {

                    var search_term = $(this).attr('value');
                    //alert(search_term); //takes what is typed in the input and alerts it
                    $.post('ajax/search.php', {search_term:search_term}, function (data) {
                        alert('Ajax Activated');
                        $('.result').html(data);
                                    $('.result li').click(function() {
                                            var result_value = $(this).text();
                                            var id_result = $(this).data("id");
                                            $('.autosuggest').attr('value', result_value);
                                            $('.id').attr('value', id_result);
                                            $('.result').html(''); 
                                    });
                    });
                });
            });

/ajax/search.php：

            require_once '../Configuration.php';
            if (isset($_POST['search_term']) == true && empty ($_POST['search_term']) == false) {

            $search_term = mysqli_real_escape_string($sql_connection, $_POST['search_term']);


            $query = mysqli_query($sql_connection,"SELECT `id`,`customer_name`,`postcode` FROM `Customers` WHERE `customer_name` LIKE '$search_term%' ");

            while(($row = mysqli_fetch_assoc($query)) !== false) {
                //loop
                  echo '<li data-id="',$row['id'],'" >',$row['customer_name'],' ',$row['postcode'],'</li>';

                }
            }


            ?>

选择客户.php：

                        $customer_id = (int) $_GET['id'];

                        $query = "SELECT * FROM Customers WHERE id = {$customer_id}";

                        $result = mysqli_query($sql_connection, $query) or die('<p>' . $query . '</p><div>' . 
                                               mysqli_error($sql_connection) . '</div>');

                        $customer = mysqli_fetch_assoc($result);  

            ?>
             <h2>Quick find users</h2>
                            <div class="block ">

                            <form action="Customer.php" method="post">


                            <input type="text" class="autosuggest"> <input type="submit" value="Go Go Go!">
                                <div class="dropdown">
                                    <ul class="result">
                                    </ul>
                                </div>
                                <input type="hidden" name="id" class="id" value="">
                            </form> 




                            </div>

问题：该字段中没有出现任何内容（这曾经有效）。

我修复了所有不起作用的路径。但它仍然没有提示搜索。

Answer 1

我被指示查看包含无限循环的 Search.php：

        while(($row = mysqli_fetch_assoc($query)) !== false) {
            //loop
              echo '<li data-id="',$row['id'],'" >',$row['customer_name'],' ',$row['postcode'],'</li>';

将其更改为：

        while($row = mysqli_fetch_assoc($query)) {
            //loop
              echo '<li data-id="',$row['id'],'" >',$row['customer_name'],' ',$row['postcode'],'</li>';