题目来源:美团。
1 题目
在外卖订单中,有时用户会指定订单的配送时间。现定义:如果用户下单日期与期望配送日期相同则认为是即时单,如果用户下单日期与期望配送时间不同则是预约单。每个用户下单时间最早的一单为用户首单,请计算用户首单中即时单的占比。
t_user_order
2 建表语句
--建表语句
CREATE TABLE t_user_order
(
order_id string COMMENT '订单ID',
user_id string COMMENT '用户ID',
order_time string comment '下单时间',
desire_date string comment '期望送达日期'
) COMMENT '用户订单记录表'
ROW FORMAT DELIMITED FIELDS TERMINATED BY '\t'
;
--插入数据
insert into t_user_order
values
('1001','001','2024-07-01 12:01:23','2024-07-01'),
('1002','001','2024-07-01 12:03:23','2024-07-02'),
('1003','002','2024-07-01 13:03:23','2024-07-02'),
('1004','002','2024-07-01 13:07:23','2024-07-01'),
('1005','003','2024-07-01 15:03:23','2024-07-01')
3 题解
- 找到用户首单,并判断是否是即时单
select order_id,
user_id,
order_time,
desire_date,
is_instant
from (select order_id,
user_id,
order_time,
desire_date,
if(to_date(order_time) = to_date(desire_date), 1, 0) as is_instant,
row_number() over (partition by user_id order by order_time asc ) as rn
from t_user_order) t
where rn = 1
执行结果
- 统计用户首单总单数和即时单数
select
count(case when is_instant = 1 then order_id end) as instant_cnt,
-- count(if(is_instant = 1, order_id, null)) as instant_cnt,
count(order_id) as total_cnt
from (select order_id,
user_id,
order_time,
desire_date,
if(to_date(order_time) = to_date(desire_date), 1, 0) as is_instant,
row_number() over (partition by user_id order by order_time asc ) as rn
from t_user_order) t
where rn = 1
执行结果
- 计算即时单比例
select
round(count(case when is_instant = 1 then order_id end)/count(order_id),2) as instant_per
from (select order_id,
user_id,
order_time,
desire_date,
if(to_date(order_time) = to_date(desire_date), 1, 0) as is_instant,
row_number() over (partition by user_id order by order_time asc ) as rn
from t_user_order) t
where rn = 1
执行结果
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