php - $stmt->fetch() 未获取数据

Question

我不明白为什么我的函数没有得到结果而只是得到空值。我正在使用 PhpStorm；我的所有连接都很好(Apache、MySQL、PhpMyAdmin)，我已经检查过它们，其他其他服务也正常工作。

这是我的dbhandler.php 文件:

public function getRating($rating_id, $user_id) {
    $stmt = $this->conn->prepare("SELECT  ur.restaurant_id, ur.service_rating, ur.food_rating, ur.music_rating FROM user_ratings ur , user u WHERE u.user_id = ? AND u.user_id = ur.user_id  AND ur.rating_id = ?");
    $stmt->bind_param("ii", $rating_id, $user_id);


    if ($stmt->execute()) {

        $stmt->bind_result( $restaurant_id, $service_rating, $food_rating, $music_rating);
        // TODO
        //$rating_id = $stmt->get_result()->fetch_assoc();
        $stmt->fetch();
        $res = array();
        $res["user_id"] = $user_id;
        $res["rating_id"] = $rating_id;
        $res["restaurant_id"] = $restaurant_id;
        $res["service_rating"] = $service_rating;
        $res["food_rating"] = $food_rating;
        $res["music_rating"] = $music_rating;
        $stmt->close();

        return $res;
    } else {
        return NULL;
    }
}

这是我的 index.php 文件

$app->get('/userRatings/:rating_id', 'authenticate', function($rating_id) {
        global $user_id;
        $response = array();
        $db = new DbHandler();

        // fetch rating
        $result = $db->getRating($rating_id, $user_id);

        if ($result != NULL) {
            $response["user_id"] = $result["user_id"];
            $response["rating_id"] = $result["rating_id"];
            $response["restaurant_id"] = $result["restaurant_id"];
            $response["service_rating"] = $result["service_rating"];
            $response["food_rating"] = $result["food_rating"];
            $response["music_rating"] = $result["music_rating"];
            echoRespnse(200, $response);
        } else {
            $response["error"] = true;
            $response["message"] = "The requested resource doesn't exists";
            echoRespnse(404, $response);
        }
    });

请求的响应是:

{"user_id":19,
"rating_id":"171",
"restaurant_id":null,
"service_rating":null,
"food_rating":null,
"music_rating":null}

Answer 1

参数值有问题。按如下方式更改参数值。

$stmt = $this->conn->prepare("SELECT  ur.restaurant_id, ur.service_rating, ur.food_rating, ur.music_rating FROM user_ratings ur , user u WHERE u.user_id = ? AND u.user_id = ur.user_id  AND ur.rating_id = ?");

$stmt->bind_param("ii", $user_id,$rating_id);

根据你的sql第一个参数应该是$user_id。不是 $ rating_id。

根据您的参数设置，没有要获取的记录。