题目描述

(通过次数8,353 | 提交次数13,192,通过率63.32%)

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。
4 号顾客的 7 号订单是首次订单,并且是即时订单。
因此,一半顾客的首次订单是即时的。
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/immediate-food-delivery-ii
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。 4 号顾客的 7 号订单是首次订单,并且是即时订单。 因此,一半顾客的首次订单是即时的。 来源:力扣(LeetCode) 链接:https://leetcode.cn/problems/immediate-food-delivery-ii
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。
4 号顾客的 7 号订单是首次订单,并且是即时订单。
因此,一半顾客的首次订单是即时的。

来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/immediate-food-delivery-ii
//测试数据
Create table If Not Exists Delivery (delivery_id int, customer_id int, order_date date, customer_pref_delivery_date date);
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');
//测试数据 Create table If Not Exists Delivery (delivery_id int, customer_id int, order_date date, customer_pref_delivery_date date); 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');
//测试数据
Create table If Not Exists Delivery (delivery_id int, customer_id int, order_date date, customer_pref_delivery_date date);

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');

解题思路

题目要求:查询所有用户的首次订单中即时订单的比例。
即时订单的含义是指下单日期与期望配送日期相同。
那么,我们可以先查询出所有用户的首次订单。然后计算出订单总数及即时订单数。最后得出即时订单的比例。

参考SQL

未特别说明的情况下,参考SQL为基于MySQL8.0实现。
select
round(
count(case when a.order_date = a.customer_pref_delivery_date then 1 else null end)*100
/count(1)
,2) immediate_percentage
from
(
select
customer_id,
order_date,
customer_pref_delivery_date,
row_number() over(partition by customer_id order by order_date) rn
from Delivery
)a
where a.rn=1;
select round( count(case when a.order_date = a.customer_pref_delivery_date then 1 else null end)*100 /count(1) ,2) immediate_percentage from ( select customer_id, order_date, customer_pref_delivery_date, row_number() over(partition by customer_id order by order_date) rn from Delivery )a where a.rn=1;
select
    round(
    count(case when a.order_date = a.customer_pref_delivery_date then 1 else null end)*100
    /count(1)
    ,2) immediate_percentage
from
(
select
    customer_id,
    order_date,
    customer_pref_delivery_date,
    row_number() over(partition by customer_id order by order_date) rn
from Delivery
)a
where a.rn=1;

picture loss