mysql GROUP_CONCAT获取分组的前几名

Posted 2020-08-09

如果是oracle 应该很容易用Partition By  实现。

比如说要获取班级的前3名，就可以用GROUP_CONCAT  + GROUP BY + substring_index实现。

考试表

DROP TABLE IF EXISTS `test`;
CREATE TABLE `test` (
`id` int(11) DEFAULT NULL,
`name` varchar(20) DEFAULT NULL,
`score` int(11) DEFAULT NULL,
`class` char(12) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8; 

插入数据

INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘1‘, ‘Bobdd‘, ‘25‘, ‘1‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘2‘, ‘xx‘, ‘20‘, ‘2‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘3‘, ‘Jack‘, ‘30‘, ‘2‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘4‘, ‘Bill‘, ‘32‘, ‘4‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘5‘, ‘Nick‘, ‘22‘, ‘3‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘6‘, ‘Kathy‘, ‘18‘, ‘3‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘7‘, ‘Steve‘, ‘36‘, ‘3‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘8‘, ‘Anne‘, ‘25‘, ‘2‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘9‘, ‘Kathy‘, ‘18‘, ‘2‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘11‘, ‘Bob1‘, ‘25‘, ‘3‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘12‘, ‘Jane1‘, ‘20‘, ‘1‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘13‘, ‘Jack1‘, ‘30‘, ‘1‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘14‘, ‘Bill1‘, ‘32‘, ‘1‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘15‘, ‘Nick1‘, ‘22‘, ‘4‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘16‘, ‘Kathy1‘, ‘18‘, ‘4‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘17‘, ‘Steve1‘, ‘36‘, ‘4‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘18‘, ‘Anne1‘, ‘25‘, ‘1‘);
INSERT INTO `test` (`id`, `name`, `score`, `class`) VALUES (‘19‘, ‘Kathy1‘, ‘18‘, ‘2‘);

 

运用group_concat + GROUP BY 分组 获取前3名

select GROUP_CONCAT(t1.id) as ids from (
SELECT t.class, substring_index(GROUP_CONCAT(t.id ORDER BY t.score desc),‘,‘,3) as id from
test t GROUP BY t.class
)t1

得到

注意 是t.id ORDER BY t.score desc 分数从高到低。

上面的语句只是获取到总的id。但是转换为列不太好弄。可以拆分用union all 来搞。

获取第一名

select GROUP_CONCAT(t1.id) as ids from (
SELECT t.class, substring_index(GROUP_CONCAT(t.id ORDER BY t.score desc),‘,‘,3) as id from
test t GROUP BY t.class 
)t1

union all

-- 第二名
SELECT t.class, substring_index(substring_index(GROUP_CONCAT(t.id ORDER BY t.score desc),‘,‘,2),‘,‘,-1) as id from
test t GROUP BY t.class

union all

-- 第三名
SELECT t.class, substring_index(substring_index(GROUP_CONCAT(t.id ORDER BY t.score desc),‘,‘,3),‘,‘,-1) as id from
test t GROUP BY t.class

好了到现在 已经获取到了一个list

用 in 来完成最后的步骤 

SELECT class,score,name FROM test where id in(
SELECT id from
(SELECT t.class, substring_index(GROUP_CONCAT(t.id ORDER BY t.score desc),‘,‘,1) as id from
test t GROUP BY t.class
union all  
SELECT t.class, substring_index(substring_index(GROUP_CONCAT(t.id ORDER BY t.score desc),‘,‘,2),‘,‘,-1) as id from
test t GROUP BY t.class
union all
SELECT t.class, substring_index(substring_index(GROUP_CONCAT(t.id ORDER BY t.score desc),‘,‘,3),‘,‘,-1) as id from
test t GROUP BY t.class) t2
) ORDER BY class asc,score desc

 最终结果