Question

我正在开店并使用输入来获得结果，现在我有调用PHP脚本的AJAX并且它调用它很好，但是我收到错误：

致命错误：未捕获的异常＆＃39; PDOException＆＃39; with message＆＃39; SQLSTATE [42000]：语法错误或访问冲突：1064

注意：错误行是$query->execute(array(':input'=>$input))行

这里是AJAX脚本（+ HTML调用函数）

                     <input type="text" name="search_item" onkeyup="showItems(this.value)" id="search_item">
                     <script>
                        function showItems(str) {
                            if (str.length == 0) { 

                            } else {
                                var xmlhttp = new XMLHttpRequest();
                                xmlhttp.onreadystatechange = function() {
                                    if (this.readyState == 4 && this.status == 200) {
                                        document.getElementById("items").innerHTML = this.responseText;
                                    }
                                };
                                xmlhttp.open("GET", "searchScript.php?iName=" + str, true);
                                xmlhttp.send();
                            }
                        }
                    </script>

这里叫做PHP：

    $input = $_REQUEST["iName"];
    $input = "%".$input."%"; 
$dsn = 'mysql:host=xxx.com;dbname=dbNameHidden;charset=utf8mb4';
$username = 'hidden';
$password = 'hidden';

try{
    // connect to mysql
    $con = new PDO($dsn,$username,$password);
    $con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
} catch (Exception $ex) {
    echo 'Not Connected '.$ex->getMessage();
}
$query = $con->prepare("SELECT * FROM store AS s INNER JOIN product_pictures AS pp ON s.product_id = pp.id INNER JOIN product_name AS pn ON s.product_id = pn.id WHERE product_name LIKE %:input% LIMIT 9 ");
$query->execute(array(':input' => $input));
$items = $query->fetchAll();

Answer 1

将通配符添加到参数：

$query = $con->prepare("SELECT ... WHERE product_name LIKE :input LIMIT 9 ");
$query->execute(array(':input' => '%' . $input. '%'));

通过这种方式，通配符包含在值中，实际上是这样的查询：

SELECT .... WHERE product_name LIKE '%name%'

Answer 2

您的查询导致LIKE %'something'%不正确。将%添加到变量而不是查询。你想要这样的东西：

$input = "%$input%";

$query = $con->prepare("SELECT * FROM store AS s 
                        INNER JOIN product_pictures AS pp ON s.product_id = pp.id
                        INNER JOIN product_name AS pn ON s.product_id = pn.id
                        WHERE product_name LIKE :input LIMIT 9 ");
$query->execute(array(':input' => $input));

致命错误：未捕获的异常＆＃39; PDOException＆＃39;消息＆＃39; SQLSTATE [42000]：语法错误或访问冲突：1064

2 个答案: