sql力扣刷题六

1164. 指定日期的产品价格

Create table If Not Exists Products (product_id int, new_price int, change_date date)
Truncate table Products
insert into Products (product_id, new_price, change_date) values ('1', '20', '2019-08-14')
insert into Products (product_id, new_price, change_date) values ('2', '50', '2019-08-14')
insert into Products (product_id, new_price, change_date) values ('1', '30', '2019-08-15')
insert into Products (product_id, new_price, change_date) values ('1', '35', '2019-08-16')
insert into Products (product_id, new_price, change_date) values ('2', '65', '2019-08-17')
insert into Products (product_id, new_price, change_date) values ('3', '20', '2019-08-18')

产品数据表: Products

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| new_price     | int     |
| change_date   | date    |
+---------------+---------+
这张表的主键是 (product_id, change_date)。
这张表的每一行分别记录了 某产品 在某个日期 更改后 的新价格。

写一段 SQL来查找在 2019-08-16 时全部产品的价格，假设所有产品在修改前的价格都是 10 。

以 任意顺序 返回结果表。

查询结果格式如下例所示。

示例 1:

输入：
Products 表:
+------------+-----------+-------------+
| product_id | new_price | change_date |
+------------+-----------+-------------+
| 1          | 20        | 2019-08-14  |
| 2          | 50        | 2019-08-14  |
| 1          | 30        | 2019-08-15  |
| 1          | 35        | 2019-08-16  |
| 2          | 65        | 2019-08-17  |
| 3          | 20        | 2019-08-18  |
+------------+-----------+-------------+
输出：
+------------+-------+
| product_id | price |
+------------+-------+
| 2          | 50    |
| 1          | 35    |
| 3          | 10    |
+------------+-------+

题解一

select p1.product_id, ifnull(p2.new_price, 10) as price
from (
    select distinct product_id
    from products
) as p1 -- 所有的产品
left join (
    select product_id, new_price 
    from products
    where (product_id, change_date) in (
        select product_id, max(change_date)
        from products
        where change_date <= '2019-08-16'
        group by product_id
    )
) as p2 -- 在 2019-08-16 之前有过修改的产品和最新的价格
on p1.product_id = p2.product_id

题解二

with temp as 
(
    select distinct product_id from Products
)

select a.product_id, ifnull(b.new_price, 10) as price from temp a
left join
(select *, rank() over(partition by product_id order by change_date DESC) as "rk" from Products where change_date<="2019-08-16") b
on a.product_id = b.product_id
where b.rk = 1 or b.rk is NULL

题解三

select
    product_id,
    price
from (
    select
        product_id,
        new_price as price,
        rank() over(partition by product_id order by change_date desc) as rk
    from Products
    where change_date <= '2019-08-16'
) t1 where rk = 1
union
select
    product_id,
    10 as price
from Products
group by product_id
having(min(change_date) > '2019-08-16')

1173. 即时食物配送 I

Create table If Not Exists Delivery (delivery_id int, customer_id int, order_date date, customer_pref_delivery_date date)
Truncate table Delivery
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('1', '1', '2019-08-01', '2019-08-02')
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('2', '5', '2019-08-02', '2019-08-02')
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('3', '1', '2019-08-11', '2019-08-11')
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('4', '3', '2019-08-24', '2019-08-26')
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('5', '4', '2019-08-21', '2019-08-22')
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('6', '2', '2019-08-11', '2019-08-13')

配送表: Delivery

+-----------------------------+---------+
| Column Name                 | Type    |
+-----------------------------+---------+
| delivery_id                 | int     |
| customer_id                 | int     |
| order_date                  | date    |
| customer_pref_delivery_date | date    |
+-----------------------------+---------+
delivery_id 是表的主键。
该表保存着顾客的食物配送信息，顾客在某个日期下了订单，并指定了一个期望的配送日期（和下单日期相同或者在那之后）。

如果顾客期望的配送日期和下单日期相同，则该订单称为 「即时订单」，否则称为「计划订单」。

写一条 SQL 查询语句获取即时订单所占的百分比， 保留两位小数。

查询结果如下所示。

示例 1:

输入：
Delivery 表:
+-------------+-------------+------------+-----------------------------+
| delivery_id | customer_id | order_date | customer_pref_delivery_date |
+-------------+-------------+------------+-----------------------------+
| 1           | 1           | 2019-08-01 | 2019-08-02                  |
| 2           | 5           | 2019-08-02 | 2019-08-02                  |
| 3           | 1           | 2019-08-11 | 2019-08-11                  |
| 4           | 3           | 2019-08-24 | 2019-08-26                  |
| 5           | 4           | 2019-08-21 | 2019-08-22                  |
| 6           | 2           | 2019-08-11 | 2019-08-13                  |
+-------------+-------------+------------+-----------------------------+
输出：
+----------------------+
| immediate_percentage |
+----------------------+
| 33.33                |
+----------------------+
解释：2 和 3 号订单为即时订单，其他的为计划订单。

题解一

select round (
    (select count(*) from Delivery where order_date = customer_pref_delivery_date) / 
    (select count(*) from Delivery) * 100,
    2
) as immediate_percentage

1174. 即时食物配送 II

Create table If Not Exists Delivery (delivery_id int, customer_id int, order_date date, customer_pref_delivery_date date)
Truncate table Delivery
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('1', '1', '2019-08-01', '2019-08-02')
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('2', '2', '2019-08-02', '2019-08-02')
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('3', '1', '2019-08-11', '2019-08-12')
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('4', '3', '2019-08-24', '2019-08-24')
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('5', '3', '2019-08-21', '2019-08-22')
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('6', '2', '2019-08-11', '2019-08-13')
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('7', '4', '2019-08-09', '2019-08-09')

配送表: Delivery

+-----------------------------+---------+
| Column Name                 | Type    |
+-----------------------------+---------+
| delivery_id                 | int     |
| customer_id                 | int     |
| order_date                  | date    |
| customer_pref_delivery_date | date    |
+-----------------------------+---------+
delivery_id 是表的主键。
该表保存着顾客的食物配送信息，顾客在某个日期下了订单，并指定了一个期望的配送日期（和下单日期相同或者在那之后）。

如果顾客期望的配送日期和下单日期相同，则该订单称为 「即时订单」，否则称为「计划订单」。

「首次订单」是顾客最早创建的订单。我们保证一个顾客只会有一个「首次订单」。

写一条 SQL 查询语句获取即时订单在所有用户的首次订单中的比例。保留两位小数。

查询结果如下所示：

Delivery 表：
+-------------+-------------+------------+-----------------------------+
| delivery_id | customer_id | order_date | customer_pref_delivery_date |
+-------------+-------------+------------+-----------------------------+
| 1           | 1           | 2019-08-01 | 2019-08-02                  |
| 2           | 2           | 2019-08-02 | 2019-08-02                  |
| 3           | 1           | 2019-08-11 | 2019-08-12                  |
| 4           | 3           | 2019-08-24 | 2019-08-24                  |
| 5           | 3           | 2019-08-21 | 2019-08-22                  |
| 6           | 2           | 2019-08-11 | 2019-08-13                  |
| 7           | 4           | 2019-08-09 | 2019-08-09                  |
+-------------+-------------+------------+-----------------------------+

Result 表：
+----------------------+
| immediate_percentage |
+----------------------+
| 50.00                |
+----------------------+
1 号顾客的 1 号订单是首次订单，并且是计划订单。
2 号顾客的 2 号订单是首次订单，并且是即时订单。
3 号顾客的 5 号订单是首次订单，并且是计划订单。