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Exercise 11:1159.Market Analysis II

1.Description

Table: Users
+----------------+---------+
| Column Name | Type |
+----------------+---------+
| user_id | int |
| join_date | date |
| favorite_brand | varchar |
+----------------+---------+user_id is the primary key of this table.
This table has the info of the users of an online shopping website where users can sell and buy items.
Table: Orders
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| order_id | int |
| order_date | date |
| item_id | int |
| buyer_id | int |
| seller_id | int |
+---------------+---------+order_id is the primary key of this table.
item_id is a foreign key to the Items table.
buyer_id and seller_id are foreign keys to the Users table.
Table: Items
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| item_id | int |
| item_brand | varchar |
+---------------+---------+item_id is the primary key of this table.
Write an SQL query to find for each user, whether the brand of the second item (by date) they sold is their favorite brand. If a user sold less than two items, report the answer for that user as no.
It is guaranteed that no seller sold more than one item on a day.
The query result format is in the following example:
Users table:
+---------+------------+----------------+
| user_id | join_date | favorite_brand |
+---------+------------+----------------+
| 1 | 2019-01-01 | Lenovo |
| 2 | 2019-02-09 | Samsung |
| 3 | 2019-01-19 | LG |
| 4 | 2019-05-21 | HP |
+---------+------------+----------------+
Orders table:
+----------+------------+---------+----------+-----------+
| order_id | order_date | item_id | buyer_id | seller_id |
+----------+------------+---------+----------+-----------+
| 1 | 2019-08-01 | 4 | 1 | 2 |
| 2 | 2019-08-02 | 2 | 1 | 3 |
| 3 | 2019-08-03 | 3 | 2 | 3 |
| 4 | 2019-08-04 | 1 | 4 | 2 |
| 5 | 2019-08-04 | 1 | 3 | 4 |
| 6 | 2019-08-05 | 2 | 2 | 4 |
+----------+------------+---------+----------+-----------+
Items table:
+---------+------------+
| item_id | item_brand |
+---------+------------+
| 1 | Samsung |
| 2 | Lenovo |
| 3 | LG |
| 4 | HP |
+---------+------------+
Result table:
+-----------+--------------------+
| seller_id | 2nd_item_fav_brand |
+-----------+--------------------+
| 1 | no |
| 2 | yes |
| 3 | yes |
| 4 | no |
+-----------+--------------------+
The answer for the user with id 1 is no because they sold nothing.
The answer for the users with id 2 and 3 is yes because the brands of their second sold items are their favorite brands.
The answer for the user with id 4 is no because the brand of their second sold item is not their favorite brand.

2.Create Table and insert into values

create table if not exists leetcode.ex_1159_user
(user_id string, join_date date,favorite_brand string) stored as orc ;
INSERT INTO table leetcode.ex_1159_user VALUES
('1' ,'2019-01-01','Lenovo'),
('2' ,'2019-02-09','Samsung'),
('3' ,'2019-01-19','LG'),
('4' ,'2019-05-21','HP')
;
create table if not exists leetcode.ex_1159_orders
(order_id string,
order_date date,
item_id string,
buyer_id string,
seller_id string
) stored as orc ;
INSERT INTO table leetcode.ex_1159_orders VALUES
('1' ,'2019-08-01' ,'4' ,'1' ,'2'),
('2 ' ,'2019-08-02' ,'2' ,'1' ,'3'),
('3 ' ,'2019-08-03' ,'3' ,'2' ,'3'),
('4' ,'2019-08-04' ,'1' ,'4' ,'2'),
('5 ' ,'2019-08-04' ,'1' ,'3' ,'4'),
('6 ' ,'2019-08-05' ,'2' ,'2' ,'4')
;
create table if not exists leetcode.ex_1159_items
(item_id string, item_brand string) stored as orc ;
INSERT INTO table leetcode.ex_1159_items VALUES
('1','Samsung'),('2','Lenovo'),('3','LG'),('4','HP') ;
Result table要返回所有user_id,不管是否有order紀錄,注意JOIN的主表
The result table should return all user_id, regardless of whether the user has an order record, pay attention to the type of JOIN