Leetcode1174. 即时食物配送 II（中等）

题目
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。

解答
先得到首次订单
还是利用分组rank的方法

select *,  @rank:=if(@pre_cid = D.customer_id, @rank +1, 1), @pre_id:=D.customer_id
from Delivery as D, (select @rank:=0, @pre_cid:=NULL) as init
order by D.customer_id, D.order_date ASC;

对于上述临时表选取rank为1的即为首次订单

select tmp.customer_id, tmp.order_date, tmp.customer_pref_delivery_date
from (select *,  @rank:=if(@pre_cid = D.customer_id, @rank +1, 1) as rank, @pre_id:=D.customer_id
from Delivery as D, (select @rank:=0, @pre_cid:=NULL) as init
order by D.customer_id, D.order_date ASC) as tmp
where tmp.rank =1;

两个date相等的订单数量除以总数量即可

select round(sum(tmp.order_date = tmp.customer_pref_delivery_date)/count(tmp.customer_id)*100,2) as immediate_percentage
from (select *,  @rank:=if(@pre_cid = D.customer_id, @rank +1, 1) as rank, @pre_id:=D.customer_id
from Delivery as D, (select @rank:=0, @pre_cid:=NULL) as init
order by D.customer_id, D.order_date ASC) as tmp
where tmp.rank =1;

其实这里首次订单分组求min即可
但这样更便于推广到第二次订单之类

别的方法
方法一、
选出每个id首次订单的日期

select customer_id,min(order_date) order_date 
from Delivery
group by customer_id

内连接选出首次订单的信息

select *
from Delivery d join (
    select customer_id,min(order_date) order_date
    from Delivery
    group by customer_id
)  w on d.customer_id=w.customer_id and d.order_date=w.order_date

首次订单中即时订单的数量

select count(*)
from Delivery d join (
    select customer_id,min(order_date) order_date
    from Delivery
    group by customer_id
)  w on d.customer_id=w.customer_id and d.order_date=w.order_date
where d.order_date=d.customer_pref_delivery_date;

除以总id数即可

select round(100*count(*)/@num_cus,2) immediate_percentage
from Delivery d join (
    select customer_id,min(order_date) order_date
    from Delivery
    group by customer_id
)  w on d.customer_id=w.customer_id and d.order_date=w.order_date
,(select @num_cus:=count(distinct customer_id) from Delivery) t
where d.order_date=d.customer_pref_delivery_date;

方法二、
如果最小的订单日期等于最小的期望配送日期则为首次订单中的即时订单

select customer_id as id, case
if(min(order_date) = min(customer_pref_delivery_date), 1, 0) as tag
from Delivery d1
group by customer_id

然后tag的和除以总数量即可


select round(100* sum(tag)/count(id),2) as immediate_percentage
from
(    
    select customer_id as id, case
    when min(order_date) = min(customer_pref_delivery_date)
    then 100.00
    else 0 end as tag
    from Delivery d1
    group by customer_id
) d2;