【发布时间】:2021-01-07 18:33:51
【问题描述】:
架构
CREATE TABLE IF NOT EXISTS `exams` (
`id` int(6) unsigned NOT NULL,
`name` varchar(100) NOT NULL,
PRIMARY KEY (`id`)
) DEFAULT CHARSET=utf8;
CREATE TABLE IF NOT EXISTS `institutions` (
`id` int(6) unsigned NOT NULL,
`name` varchar(100) NOT NULL,
PRIMARY KEY (`id`)
) DEFAULT CHARSET=utf8;
CREATE TABLE IF NOT EXISTS `exam_scores` (
`id` int(6) unsigned NOT NULL,
`exam_id` int(6) NOT NULL,
`institution_id` int(6) NOT NULL,
`score` int(5)
PRIMARY KEY (`id`)
) DEFAULT CHARSET=utf8;
INSERT INTO `exams` (`id`, `name`) VALUES
('1', 'exam1'),
('2', 'exam2'),
('3', 'exam3'),
('4', 'exam4');
('5', 'exam5');
INSERT INTO `institutions` (`id`, `name`) VALUES
('1', 'institution1'),
('2', 'institution2'),
('3', 'institution3'),
('4', 'institution4');
('5', 'institution5');
INSERT INTO `exam_scores` (`id`, `exam_id`, `institution_id`, `score`) VALUES
('1', '1', 1, 40),
('2', '2', 1, 45),
('3', '3', 2, 35),
('4', '1', 2, 30);
('5', '4', 3, 40);
现在用户将输入exm1
我正在尝试创建一个查询来查找所有相关考试,如下所示。
查找匹配输入 exm1 的考试,还可以找到匹配机构 inn exam_scores 表中存在的其他考试。
示例1:输入exm4
desired output
| exm4 |
示例2:输入exm3
desired ouput
| exm3 |
| exm1 |
示例3:输入exm1
desired output
| exm1 |
| exm2 |
| exm3 |
到目前为止,我只提出了一个只提供匹配考试的查询:)
select exams.name from exams
inner join exam_scores on exam_scores.exam_id = exams.id
// ??
where exams.id = 1
【问题讨论】:
-
exm1的输出不应该是exm1和exm2? -
@Tajni 因为
exm1匹配institution (id: 2)和institution id 2有exm3 :) 所以输出是正确的。
标签: mysql sql join inner-join where-clause