Question

我正在尝试使用php和AJAX创建一个搜索框。但是，当我在搜索框中键入数据库中已有的名称时，不会显示任何建议。什么都没发生。

我测试了php并且似乎正在工作。我错过了什么？

使用PDO进行代码更新

require_once('connect_pdo.php');

$searchTerm = $_GET['term'];

$stmtus = $conn->prepare("SELECT * FROM `Schools` WHERE `School` LIKE :School");
$stmtus->bindValue(':School', '%' . $searchTerm . '%');
$searchTerm = $_POST['school'];;
$stmtus -> execute();


while ($result = $stmtus->fetch(PDO::FETCH_ASSOC)) {

$school = $result['School'];

}
echo json_encode($school);

AJAX

 <script type="text/javascript"
    src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js"></script>
    <script type="text/javascript"
    src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.8.18/jquery-ui.min.js"></script>
    <link rel="stylesheet" type="text/css"
    href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.8/themes/base/jquery-ui.css" />

    <script type="text/javascript">
            $(document).ready(function(){
                $("#school").autocomplete({
                    source:'schoolfind.php',
                    minLength:3
                });
            });
    </script>

HTML

<label>What school does the child attend<input  type="text" name="school" id="school" /></label>

Answer 1

您使用的任何绑定锚点（例如:School）都需要在查询字符串中显示未加引号。即，不要这样做：

$stmtus = $conn->prepare("SELECT * FROM `Schools` WHERE `School` LIKE '%".$searchTerm."%'");
$stmtus->bindParam(':School', $searchTerm);

这样做：

$stmtus = $conn->prepare("SELECT * FROM `Schools` WHERE `School` LIKE :School");
$stmtus->bindValue(':School', '%' . $searchTerm . '%');

AJAX / PHP搜索不起作用

1 个答案: