use*_*823 0 php mysql database
这是我试图运行它的查询工作,但现在有一个原因我无法弄清楚它现在放弃了.
SELECT films.movie_title, films.rating, films.actor, reviewed.review
FROM films
INNER JOIN reviewed
ON films.movie_id=reviewed.review_id
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经过这些表格后,我遇到了这个问题.认为这可能是问题所在.
ALTER TABLE `reviewed`
ADD CONSTRAINT `reviewed_ibfk_1` FOREIGN KEY (`movie_id`) REFERENCES `films` (`movie_id`),
ADD CONSTRAINT `reviewed_ibfk_2` FOREIGN KEY (`movie_id`) REFERENCES `films` (`movie_id`) ON DELETE CASCADE;
--
-- Table structure for table `films`
--
CREATE TABLE IF NOT EXISTS `films` (
`movie_id` int(4) NOT NULL AUTO_INCREMENT,
`movie_title` varchar(100) NOT NULL,
`actor` varchar(100) NOT NULL,
`rating` varchar(20) NOT NULL,
PRIMARY KEY (`movie_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=23 ;
--
-- Dumping data for table `films`
--
INSERT INTO `films` (`movie_id`, `movie_title`, `actor`, `rating`) VALUES
(22, 'Batman Begins', 'Bale', 'Good');
-- --------------------------------------------------------
--
-- Table structure for table `reviewed`
--
CREATE TABLE IF NOT EXISTS `reviewed` (
`review_id` int(4) NOT NULL AUTO_INCREMENT,
`review` mediumtext NOT NULL,
`movie_id` int(4) NOT NULL,
PRIMARY KEY (`review_id`),
KEY `movie_id` (`movie_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ;
--
-- Dumping data for table `reviewed`
--
INSERT INTO `reviewed` (`review_id`, `review`, `movie_id`) VALUES
(1, 'Really good film.', 22);
--
ALTER TABLE `reviewed`
ADD CONSTRAINT `reviewed_ibfk_1` FOREIGN KEY (`movie_id`) REFERENCES `films` (`movie_id`),
ADD CONSTRAINT `reviewed_ibfk_2` FOREIGN KEY (`movie_id`) REFERENCES `films` (`movie_id`) ON DELETE CASCADE;
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小智 5
看起来你在两个主键连接(films.movie_id = reviewed.review_id),而不是一个小学和一个外国(films.movie_id = reviewed.movie_id).
尝试以下内容:
SELECT films.movie_title, films.rating, films.actor, reviewed.review
FROM films
INNER JOIN reviewed
ON films.movie_id=reviewed.movie_id
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