SQL273. 牛客的课程订单分析(三)
描述
有很多同学在牛客购买课程来学习，购买会产生订单存到数据库里。
有一个订单信息表(order_info)，简况如下:
| id | user_id | product_name | status | client_id | date |
| 1 | 557336 | C++ | no_completed | 1 | 2025-10-10 |
| 2 | 230173543 | Python | completed | 2 | 2025-10-12 |
| 3 | 57 | JS | completed | 3 | 2025-10-23 |
| 4 | 57 | C++ | completed | 3 | 2025-10-23 |
| 5 | 557336 | Java | completed | 1 | 2025-10-23 |
| 6 | 57 | Java | completed | 1 | 2025-10-24 |
| 7 | 557336 | C++ | completed | 1 | 2025-10-25 |
第1行表示user_id为557336的用户在2025-10-10的时候使用了client_id为1的客户端下了C++课程的订单，但是状态为没有购买成功。
第2行表示user_id为230173543的用户在2025-10-12的时候使用了client_id为2的客户端下了Python课程的订单，状态为购买成功。
。。。
最后1行表示user_id为557336的用户在2025-10-25的时候使用了client_id为1的客户端下了C++课程的订单，状态为购买成功。
请你写出一个sql语句查询在2025-10-15以后，同一个用户下单2个以及2个以上状态为购买成功的C++课程或Java课程或Python课程的订单信息，并且按照order_info的id升序排序，以上例子查询结果如下:
| id | user_id | product_name | status | client_id | date |
| 4 | 57 | C++ | completed | 3 | 2025-10-23 |
| 5 | 557336 | Java | completed | 1 | 2025-10-23 |
| 6 | 57 | Java | completed | 1 | 2025-10-24 |
| 7 | 557336 | C++ | completed | 1 | 2025-10-25 |
解析:
id为4，6的订单满足以上条件，输出它们的对应的信息;
id为5，7的订单满足以上条件，输出它们的对应的信息;
按照id升序排序
示例1
输入:
drop table if exists order_info; CREATE TABLE order_info ( id int(4) NOT NULL, user_id int(11) NOT NULL, product_name varchar(256) NOT NULL, status varchar(32) NOT NULL, client_id int(4) NOT NULL, date date NOT NULL, PRIMARY KEY (id)); INSERT INTO order_info VALUES (1,557336,'C++','no_completed',1,'2025-10-10'), (2,230173543,'Python','completed',2,'2025-10-12'), (3,57,'JS','completed',3,'2025-10-23'), (4,57,'C++','completed',3,'2025-10-23'), (5,557336,'Java','completed',1,'2025-10-23'), (6,57,'Java','completed',1,'2025-10-24'), (7,557336,'C++','completed',1,'2025-10-25');
输出:
4|57|C++|completed|3|2025-10-23 5|557336|Java|completed|1|2025-10-23 6|57|Java|completed|1|2025-10-24 7|557336|C++|completed|1|2025-10-25
Sqlite 解法, 执行用时: 10ms, 内存消耗: 3584KB, 提交时间: 2022-01-26
select id, user_id, product_name, status, client_id, date from
(select *, count() over(partition by user_id) num from order_info
where product_name in ('C++', 'Java', 'Python')
and status = 'completed' and date > '2025-10-15')
where num >= 2
order by id
Sqlite 解法, 执行用时: 10ms, 内存消耗: 3624KB, 提交时间: 2021-09-07
select *
from order_info
where user_id in (
select user_id
from order_info
where date>'2025-10-15'
and status=='completed'
and product_name in ('C++','Java','Python')
group by user_id
having count(*)>=2
)
and date>'2025-10-15'
and status=='completed'
and product_name in ('C++','Java','Python')
order by id
Sqlite 解法, 执行用时: 10ms, 内存消耗: 3672KB, 提交时间: 2021-08-07
select * from order_info
where date>'2025-10-15'
and product_name in ("C++","Java","Python")
and status="completed"
and user_id in
(
select user_id from order_info
where date>'2025-10-15' and status='completed'
and product_name in('C++','Java','Python')
group by user_id
having count(1) >1
)
order by id asc
Sqlite 解法, 执行用时: 10ms, 内存消耗: 3716KB, 提交时间: 2021-09-09
with b as (
select user_id from order_info
where date >'2025-10-15' and status = 'completed' and product_name in ('C++','Java','Python')
group by user_id
having count (id) >= 2
order by id
)
select * from order_info
where date >'2025-10-15' and status = 'completed' and product_name in ('C++','Java','Python')
and user_id in b
order by id ;
Sqlite 解法, 执行用时: 10ms, 内存消耗: 3756KB, 提交时间: 2021-09-08
select o1.*
from order_info o1,(
select user_id
from order_info
where date > '2025-10-15' and status ='completed' and product_name in('C++','Java','Python')
group by user_id
having count(*)>=2 ) o2
where o1.user_id=o2.user_id and date > '2025-10-15' and status ='completed' and product_name in('C++','Java','Python')
order by id