從與其他表格重合的表格中選擇數據

Question

如何顯示電影的評論？

這是我的查詢 -

$query = "SELECT movies.movie_id, movies.movie_title, 
        FROM movies 
        INNER JOIN reviews 
        ON reviews.review_id=reviews.fk_movie_review"; 




CREATE TABLE IF NOT EXISTS `movies` (
    `movie_id` int(4) NOT NULL AUTO_INCREMENT, 
    `movie_title` varchar(100) NOT NULL, 
    `actor` varchar(100) NOT NULL, 
    `fk_movie_ratings` int(2) NOT NULL, 
    PRIMARY KEY (`movie_id`) 
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ; 

-- 


INSERT INTO `movies` (`movie_id`, `movie_title`, `actor`, `fk_movie_ratings`) VALUES 
(1, 'Batman Begins', 'Christian Bale', 5); 



CREATE TABLE IF NOT EXISTS `ratings` (
    `rating_id` int(2) NOT NULL, 
    `rating_name` varchar(50) NOT NULL, 
    PRIMARY KEY (`rating_id`) 
) ENGINE=InnoDB DEFAULT CHARSET=latin1; 



INSERT INTO `ratings` (`rating_id`, `rating_name`) VALUES 
(1, 'Excellent'), 
(2, 'Brilliant'), 
(3, 'Good'), 
(4, 'Ok'), 
(5, 'Terrible'); 



-- 
-- Table structure for table `reviews` 
-- 

CREATE TABLE IF NOT EXISTS `reviews` (
    `review_id` int(4) NOT NULL AUTO_INCREMENT, 
    `movie_review` mediumtext NOT NULL, 
    `fk_movie_review` int(4) NOT NULL, 
    PRIMARY KEY (`review_id`) 
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ; 



INSERT INTO `reviews` (`review_id`, `movie_review`, `fk_movie_review`) VALUES 
(1, 'Wicked', 1); 

Answer 1

$query = "SELECT movies.movie_id, movies.movie_title, reviews.* 
        FROM movies 
        INNER JOIN reviews 
        ON movies.movie_id=reviews.fk_movie_review"; 

加盟條件並非指電影臺？

而且你還需要更多的列在選擇托馬斯·安德魯斯發現

Answer 2

有一個問題你加入，你已經從review表ON子句中指定的兩列，你還沒有加入電影臺

SELECT movies.movie_id, movies.movie_title,reviews.movie_review 
FROM movies 
INNER JOIN reviews 
ON movies.movie_id=reviews.fk_movie_review 

Answer 3

試試這個：

SELECT MOVIES.MOVIE_ID, 
     MOVIES.MOVIE_TITLE, 
     REVIEWS.MOVIE_REVIEW 
FROM MOVIES 
     LEFT JOIN REVIEWS 
       ON REVIEWS.FK_MOVIE_REVIEW = MOVIES.MOVIE_ID