Question

我目前正在尝试执行mysqli_query，但是当我运行它时，我被告知查询失败了。我不确定是什么问题，因为我的php文件正在连接到数据库服务器。以下是我的代码：

<html>
<head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1">

<link rel="icon" href="../img/favicon.ico">
<link href="css/custom.css" rel="stylesheet" type="text/css" />

<title>Assignment 8: Building a "LAMP" Website for your MySQL Data</title>
</head>
<body>
<div id="container">
    <header>
        <h1><mark>"LAMP" Website for MYSQL Movie Databse<mark></h1>
    </header>

    <section>
    <p>Please include the following notes on your page along with an introduction to the data for your user: 1. The source for your data. Is this meant to be a user-based system? or will the data come from downloads? Are the current data test data only? 2. Write at least one paragraph to describe your project. 3. Write at least one paragraph to describe your project. You might wish to refer to our class discussion on specifications for ideas.</p>

        <?php
            /* Connecting and selecting database */
            include ('/home/as7695/db_files/db_config.php');

            $db_link = new mysqli($db_server, $db_user, $db_password, $db_name);
            if ($db_link->connect_errno) {
            print( "Failed to connect to MySQL: (" .$db_link->connect_errno . ") ".$db_link->connect_error);
            }
            print("<p>Connection: ".$db_link->host_info . "<br />\n");
            print( nl2br("Connected successfully\n"));

            /* Setting variables */
            $nationality = $_POST['nationality'];
            $genre = $_POST['genre'];
            $limit = $_POST['limit'];
            $sortkey = $_POST ['sortkey'];

            /* Performing SQL query using criteria supplied by the user on the HTML form */
            $query = "SELECT CONCAT (movies.title, ' (', movies.release_year, ')') AS Title, CONCAT (movies.rating) AS Rating, CONCAT (directors.first_name, ' ', directors.last_name) AS Director, GROUP CONCAT (actors.first_name, ' ', actors.last_name separator ', ') AS Actors, CONCAT (movies.description) AS Description
            FROM movies
            INNER JOIN directors ON movies.directorID = directors.directorID
            INNER JOIN movie_actors ON movies.movieID = movie_actors.movieID
            INNER JOIN actors ON movie_actors.actorID = actors.actorID
            INNER JOIN movie_genres ON movies.movieID = movie_genres.movieID
            INNER JOIN genres ON movie_genres.genreID = genres.genreID
            WHERE directors.nationality = '$nationality' AND genres.genre_name = '$genre'
            GROUP BY movies.movieID
            ORDER BY movies.release_year '$sortkey'
            LIMIT '$limit'";

            /* Store result as a variable and see how many records are returned */
            $result = mysqli_query($db_link,$query) or die("Query failed : " . mysqli_error());

            /* Closing connection */
            mysqli_close($db_link);
        ?>
    </section>
</div>
</body>
</html>

mysqli_query将不会执行

0 个答案: