Mysql 多对多关系查询统计@TOC
有三张表:
- user(用户表,sex=0表示男生,1表示女生)
- ach(成就表)
- user_ach(关系表)
每个用户可以获得多个成就,每个成就对应多个用户。用一条sql查询出每个成就下男生和女生的人数。
user表:
CREATE TABLE `user` (
`id` int(11) NOT NULL,
`name` varchar(45) DEFAULT NULL,
`sex` tinyint(1) DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4';
ach表:
CREATE TABLE `ach` (
`id` int(11) NOT NULL,
`ach_name` varchar(45) DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4';
user_ach表:
CREATE TABLE `user_ach` (
`id` int(11) NOT NULL,
`user_id` int(11) DEFAULT NULL,
`ach_id` int(11) DEFAULT NULL,
PRIMARY KEY (`id`),
KEY `fk_user_idx` (`user_id`),
KEY `fk_ach_idx` (`ach_id`),
CONSTRAINT `fk_ach` FOREIGN KEY (`ach_id`) REFERENCES `ach` (`id`),
CONSTRAINT `fk_user` FOREIGN KEY (`user_id`) REFERENCES `user` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
查询语句:
select
t.ach_name,
max(case t.sex when 0 then t.num else 0 end) "男生",
max(case t.sex when 1 then t.num else 0 end) "女生"
from
(
select b.ach_name ach_name,a.sex sex,count(c.ach_id) num
from user a,ach b,user_ach c
where a.id = c.user_id and b.id = c.ach_id
group by c.ach_id,a.sex) t
group by t.ach_name
order by ach_name;
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