题目
Table: Users
+----------------+---------+
| Column Name | Type |
+----------------+---------+
| user_id | int |
| join_date | date |
| favorite_brand | varchar |
+----------------+---------+
此表主键是 user_id，表中描述了购物网站的用户信息，用户可以在此网站上进行商品买卖。
Table: Orders
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| order_id | int |
| order_date | date |
| item_id | int |
| buyer_id | int |
| seller_id | int |
+---------------+---------+
此表主键是 order_id，外键是 item_id 和（buyer_id，seller_id）。
Table: Item
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| item_id | int |
| item_brand | varchar |
+---------------+---------+
此表主键是 item_id。
请写出一条SQL语句以查询每个用户的注册日期和在 2019 年作为买家的订单总数。
查询结果格式如下：
Users table:
+---------+------------+----------------+
| user_id | join_date | favorite_brand |
+---------+------------+----------------+
| 1 | 2018-01-01 | Lenovo |
| 2 | 2018-02-09 | Samsung |
| 3 | 2018-01-19 | LG |
| 4 | 2018-05-21 | HP |
+---------+------------+----------------+
Orders table:
+----------+------------+---------+----------+-----------+
| order_id | order_date | item_id | buyer_id | seller_id |
+----------+------------+---------+----------+-----------+
| 1 | 2019-08-01 | 4 | 1 | 2 |
| 2 | 2018-08-02 | 2 | 1 | 3 |
| 3 | 2019-08-03 | 3 | 2 | 3 |
| 4 | 2018-08-04 | 1 | 4 | 2 |
| 5 | 2018-08-04 | 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:
+-----------+------------+----------------+
| buyer_id | join_date | orders_in_2019 |
+-----------+------------+----------------+
| 1 | 2018-01-01 | 1 |
| 2 | 2018-02-09 | 2 |
| 3 | 2018-01-19 | 0 |
| 4 | 2018-05-21 | 0 |
+-----------+------------+----------------+
生成数据
CREATE TABLE Users1 (user_id INT, join_date DATE, favorite_brand VARCHAR(20))
CREATE TABLE Orders4 (order_id INT, order_date DATE, item_id INT, buyer_id INT, seller_id INT)
CREATE TABLE Items (item_id INT, item_brand VARCHAR(20))
INSERT INTO Users1 (user_id, join_date, favorite_brand) VALUES ('1', '2018-01-01', 'Lenovo');
INSERT INTO Users1 (user_id, join_date, favorite_brand) VALUES ('2', '2018-02-09', 'Samsung');
INSERT INTO Users1 (user_id, join_date, favorite_brand) VALUES ('3', '2018-01-19', 'LG');
INSERT INTO Users1 (user_id, join_date, favorite_brand) VALUES ('4', '2018-05-21', 'HP');
INSERT INTO Orders4 (order_id, order_date, item_id, buyer_id, seller_id) VALUES ('1', '2019-08-01', '4', '1', '2');
INSERT INTO Orders4 (order_id, order_date, item_id, buyer_id, seller_id) VALUES ('2', '2018-08-02', '2', '1', '3');
INSERT INTO Orders4 (order_id, order_date, item_id, buyer_id, seller_id) VALUES ('3', '2019-08-03', '3', '2', '3');
INSERT INTO Orders4 (order_id, order_date, item_id, buyer_id, seller_id) VALUES ('4', '2018-08-04', '1', '4', '2');
INSERT INTO Orders4 (order_id, order_date, item_id, buyer_id, seller_id) VALUES ('5', '2018-08-04', '1', '3', '4');
INSERT INTO Orders4 (order_id, order_date, item_id, buyer_id, seller_id) VALUES ('6', '2019-08-05', '2', '2', '4');
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');
解答
查询每个用户的注册日期和在 2019 年作为买家的订单总数。
先查询2019 年作为买家的订单总数
SELECT O.`buyer_id`, COUNT(O.buyer_id) AS cnt_2019
FROM Orders4 AS O
WHERE O.`order_date` BETWEEN '2019-01-01' AND '2019-12-31'
GROUP BY O.`buyer_id`;
与user表进行连接即可
SELECT U.`user_id`, U.`join_date`, IFNULL(tmp.cnt_2019, 0) AS orders_in_2019
FROM Users1 AS U
LEFT JOIN (SELECT O.`buyer_id`, COUNT(O.buyer_id) AS cnt_2019
FROM Orders4 AS O
WHERE O.`order_date` BETWEEN '2019-01-01' AND '2019-12-31'
GROUP BY O.`buyer_id`) AS tmp
ON U.`user_id` = tmp.buyer_id;
选2019年的字段可以用 year(order_date)='2019'
select u.user_id buyer_id, u.join_date, ifnull(t.cnt,0) orders_in_2019
from Users u
left join
(select buyer_id, count(distinct order_id) cnt
from orders
where year(order_date)='2019'
group by buyer_id) t
on u.user_id=t.buyer_id;
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版权声明：本文为CSDN博主「红楼终究一场梦」的原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接及本声明。
原文链接：https://blog.csdn.net/Hello_JavaScript/java/article/details/104610178