单挑力扣（LeetCode）SQL笔试题：1173. 即时食物配送 I（难度：简单）

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。


来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/immediate-food-delivery-i

#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', '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');

select
    case when count(1)>0 then round(count(case when order_date = customer_pref_delivery_date then 1 else null end)*100/count(1),2) else 0.00 end immediate_percentage
from Delivery;