php prepare error

问题描述

问 题

php mysqli prepare 错误，不知道什么原因
代码如下：

            $mysqli = new mysqli($mysql_server_name, $mysql_username, $mysql_password);
            if (mysqli_connect_errno()) {
                die(sprintf("Connect failed: %s\n", mysqli_connect_error()));
            }
                $sql = "SELECT name FROM table1 WHERE name=? OR first=?";

            if ($stmt = $mysqli->prepare($sql)) {
                    $stmt->bind_param("ss", 'zzz','zzz');

                $stmt->execute();

                while ($row = $stmt->fetch()) {
                    $categories[] = $row;
                }
                $json=json_encode($categories);
                $mysqli->close();
                echo $json;

            }else{$mysqli->close();die('Prepare error');}
            
?>

解决方案

用的哪个database呀。

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