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;