致命错误:在第24行的demo.php中调用未定义的方法mysqli_stmt :: get_result()

时间:2016-03-03 10:54:41

标签: php

也许这个问题是重复的。但我找不到我的问题的答案。 我遇到了这个问题Fatal error: Call to undefined method mysqli_stmt::get_result() in demo.php on line 24 第24行位于代码$tasks = $stmt->get_result();

的这一部分

这是我的demo.php代码。

<?php

/**
 * Class to handle all db operations
 * This class will have CRUD methods for database tables
 *
 * @author Ravi Tamada
 * @link URL Tutorial link
 */
class Demo {

    private $conn;

    function __construct() {
        require_once dirname(__FILE__) . '/include/db_connect.php';
        // opening db connection
        $db = new DbConnect();
        $this->conn = $db->connect();
    }

    public function getAllChatRooms() {
        $stmt = $this->conn->prepare("SELECT * FROM chat_rooms");
        $stmt->execute();
        $tasks = $stmt->get_result();
        $stmt->close();
        return $tasks;
    }

    public function getAllUsers() {
        $stmt = $this->conn->prepare("SELECT * FROM users");
        $stmt->execute();
        $tasks = $stmt->get_result();
        $stmt->close();
        return $tasks;
    }

    public function getDemoUser() {
        $name = 'AndroidHive';
        $email = 'admin@androidhive.info';

        $stmt = $this->conn->prepare("SELECT user_id from users WHERE email = ?");
        $stmt->bind_param("s", $email);
        $stmt->execute();
        $stmt->store_result();
        $num_rows = $stmt->num_rows;
        if ($num_rows > 0) {
            $stmt->bind_result($user_id);
            $stmt->fetch();
            return $user_id;
        } else {
            $stmt = $this->conn->prepare("INSERT INTO users(name, email) values(?, ?)");
            $stmt->bind_param("ss", $name, $email);
            $result = $stmt->execute();
            $user_id = $stmt->insert_id;
            $stmt->close();
            return $user_id;
        }
    }

}

?>

在我的index.php中,我有代码来调用该函数。 这是示例代码

<?php
$chatrooms = $demo->getAllChatRooms();
foreach ($chatrooms as $key => $chatroom) {
    $cls = $key == 0 ? 'selected' : '';
    ?>
    <li id="<?= $chatroom['chat_room_id'] ?>" class="<?= $cls ?>">
        <label><?= $chatroom['name'] ?></label>
        <span>topic_<?= $chatroom['chat_room_id'] ?></span>
    </li>
    <?php
}
?>

请帮我构建此代码。

谢谢。

1 个答案:

答案 0 :(得分:1)

您需要安装mysqlnd支持。 (MySQL本机驱动程序)