LeetCode 1159.市场分析2

数据准备

Create table If Not Exists Users (user_id int, join_date date, favorite_brand varchar(10));
Create table If Not Exists Orders (order_id int, order_date date, item_id int, buyer_id int, seller_id int);
Create table If Not Exists Items (item_id int, item_brand varchar(10));
Truncate table Users;
insert into Users (user_id, join_date, favorite_brand) values ('1', '2019-01-01', 'Lenovo');
insert into Users (user_id, join_date, favorite_brand) values ('2', '2019-02-09', 'Samsung');
insert into Users (user_id, join_date, favorite_brand) values ('3', '2019-01-19', 'LG');
insert into Users (user_id, join_date, favorite_brand) values ('4', '2019-05-21', 'HP');
Truncate table Orders;
insert into Orders (order_id, order_date, item_id, buyer_id, seller_id) values ('1', '2019-08-01', '4', '1', '2');
insert into Orders (order_id, order_date, item_id, buyer_id, seller_id) values ('2', '2019-08-02', '2', '1', '3');
insert into Orders (order_id, order_date, item_id, buyer_id, seller_id) values ('3', '2019-08-03', '3', '2', '3');
insert into Orders (order_id, order_date, item_id, buyer_id, seller_id) values ('4', '2019-08-04', '1', '4', '2');
insert into Orders (order_id, order_date, item_id, buyer_id, seller_id) values ('5', '2019-08-04', '1', '3', '4');
insert into Orders (order_id, order_date, item_id, buyer_id, seller_id) values ('6', '2019-08-05', '2', '2', '4');
Truncate table Items;
insert into Items (item_id, item_brand) values ('1', 'Samsung');
insert into Items (item_id, item_brand) values ('2', 'Lenovo');
insert into Items (item_id, item_brand) values ('3', 'LG');
insert into Items (item_id, item_brand) values ('4', 'HP');

需求

写一个 SQL 查询确定每一个用户按日期顺序卖出的第二件商品的品牌是否是他们最喜爱的品牌。如果一个用户卖出少于两件商品，查询的结果是 no 。

输入

输出

with t1 as (
    select u.*,o.*,I.item_brand
from Users as u
    left join Orders as o on u.user_id=o.seller_id
    left join Items I on o.item_id = I.item_id
),t2 as (
    select user_id,favorite_brand,item_brand,
       row_number() over (partition by user_id order by order_date) as rn1
    from t1
),t3 as (
    select user_id,
       case
           when rn1=1 then 'no'
           when (rn1=2 and t2.favorite_brand=t2.item_brand) then 'yes'
           else 'no'
       end as 2nd_item_fav_brand
    from t2
)
select user_id as seller_id,
       max(2nd_item_fav_brand) as 2nd_item_fav_brand
from t3
group by user_id
;