我有一个用户表,看起来像这样:
ID NAME
1 Robin
2 Edward
3 Donald
4 Julie
第二张表 user_tree
ID USER_ID TREE
1 1 ["2","3","4"]
2 2 ["3","4"]
3 3 ["4"]
4 4 []
第3张表订单
ID AMOUNT USER_ID
1 150 2
2 300 3
3 200 4
4 500 3
预期结果,使用SQL查询获取下表(使用select ..创建表mlm_team)
ID USER_ID TEAM_SUM
1 1 1150
2 2 1000
3 3 200
是否有SQL专家可以帮助我?
答案 0 :(得分:0)
按照以下步骤操作:
- Create table and insert record.
- Get JSON type for array value
- Use "JSON_CONTAINS" for find MLM chain
在这里解释:
create table users(id int, name varchar(50));
create table user_tree(id int, user_id int, tree json );
create table orders(id int, amount int, user_id int);
insert into users values(1, 'Robin');
insert into users values(2, 'Edward');
insert into users values(3 , 'Donald');
insert into users values(4 , 'Julie');
select * from users;
insert into user_tree values(1, 1 , '["2","3","4"]');
insert into user_tree values(2 , 2 , '["3","4"]');
insert into user_tree values(3 , 3 , '["4"]');
insert into user_tree values(4 , 4 , '[]');
select * from user_tree;
insert into orders values(1 , 150 , 2);
insert into orders values(2 , 300 , 3);
insert into orders values(3 , 200 , 4);
insert into orders values(4 , 500 , 3);
select * from orders;
- 此处的最终查询:
SELECT u1.id, u1.user_id as user_id, sum(amount)
FROM user_tree u1
left outer join user_tree u2 on JSON_CONTAINS(u1.tree, concat('["', u2.user_id ,'"]'))
left outer join orders o1 on o1.user_id = u2.user_id
where u2.user_id
group by u1.id, u1.user_id
order by u1.id;
从此处LINK查找SQL小提琴详细信息。