深入解析MySQL的窗口函数

函数名(字段名) over(子句) 

函数名(字段名) over(partition by <要分列的组> order by <要排序的列> rows between <数据范围>) 

rows between 2 preceding and current row # 取本行和前面两行
rows between unbounded preceding and current row # 取本行和之前所有的行 
rows between current row and unbounded following # 取本行和之后所有的行 
rows between 3 preceding and 1 following # 从前面三行和下面一行，总共五行 
# 当order by后面没有rows between时，窗口规范默认是取本行和之前所有的行
# 当order by和rows between都没有时，窗口规范默认是分组下所有行（rows between unbounded preceding and unbounded following） 

-- 现有2018~2020某电商平台订单信息表user_trade
create table user_trade (
 user_name  varchar(20) COMMENT '用户名',
 piece  int COMMENT '购买数量',
 price  double COMMENT '价格',
 pay_amount double COMMENT '支付金额',
 goods_category varchar(20) COMMENT '商品品类',
 pay_time date COMMENT '支付日期'
);

-- 需求1: 查询出2019年每月的支付总额和当年累积支付总额 
select a.mon,a.pay_amount,sum(a.pay_amount) over(order by a.mon) as sum_amount
from(
select month(a.pay_time) as mon,sum(a.pay_amount) as pay_amount
from user_trade a
where year(a.pay_time) = '2019'
group by month(a.pay_time)
) a ;
-- 需求2:查询出2018-2019年每月的支付总额和当年累积支付总额
select a.*,sum(a.pay_amount) over(partition by a.year order by a.mon) as sum_amount
from(
select year(a.pay_time) as year,month(a.pay_time) as mon,sum(a.pay_amount) as pay_amount
from user_trade a
where year(a.pay_time) in('2018','2019')
group by year(a.pay_time),month(a.pay_time)
) a ;

-- 需求3: 查询出2019年每个月的近三月移动平均支付金额
select a.mon,a.pay_amount,
avg(a.pay_amount) over(order by a.mon rows between 2 preceding and current row) as avg_amount
from(
select month(a.pay_time) as mon,sum(a.pay_amount) as pay_amount
from user_trade a
where year(a.pay_time) = '2019'
group by month(a.pay_time)
) a ;

-- 需求4: 查询出每四个月的最大月总支付金额
select 
a.mon,
a.pay_amount,
max(a.pay_amount) over(order by a.mon rows between 3 preceding and current row) as max_amount
from(
select SUBSTRING(a.pay_time,1,7) as mon,sum(a.pay_amount) as pay_amount
from user_trade a
group by SUBSTRING(a.pay_time,1,7)
)a ;

-- 需求4: 查询出每四个月的最大月总支付金额
select 
a.mon,
a.pay_amount,
max(a.pay_amount) over(order by a.mon rows between 3 preceding and current row) as max_amount
from(
select SUBSTRING(a.pay_time,1,7) as mon,sum(a.pay_amount) as pay_amount
from user_trade a
group by SUBSTRING(a.pay_time,1,7)
)a ;

-- 需求6: 查询出将2020年2月的支付用户，按照支付金额分成5组后的结果
select 
a.user_name,
sum(a.pay_amount) as pay_amount,
ntile(5) over(order by sum(a.pay_amount) desc) as level
from user_trade a
where SUBSTRING(a.pay_time,1,7) = '2020-02'
group by a.user_name;
-- 需求7: 查询出2020年支付金额排名前30%的所有用户
select a.user_name,a.pay_amount
from (
select 
a.user_name,
sum(a.pay_amount) as pay_amount,
ntile(10) over(order by sum(a.pay_amount) desc) as level
from user_trade a
where year(a.pay_time) = '2020'
group by a.user_name
) a 
where a.level in(1,2,3);

-- 需求8: 查询出King和West的时间偏移(前N行)
select a.user_name,a.pay_time,
lag(a.pay_time,1,a.pay_time) over(partition by a.user_name order by a.pay_time) as lag1,
-- 没有传入偏移量，那么默认就是1，找不到的话，此处也没有给默认值，为null
lag(a.pay_time) over(partition by a.user_name order by a.pay_time) as lag2,
lag(a.pay_time,2,a.pay_time) over(partition by a.user_name order by a.pay_time) as lag3,
lag(a.pay_time,2) over(partition by a.user_name order by a.pay_time) as lag4
from user_trade a 
where a.user_name in('King','West');

-- 需求9: 查询出支付时间间隔超过100天的用户数
select count(distinct a.user_name)
from (
select a.user_name,a.pay_time,
lag(a.pay_time) over(partition by a.user_name order by a.pay_time) as lg
from user_trade a 
) a 
where DATEDIFF(a.pay_time,a.lg) >100;
# 需求9运行结果为180
-- 需求10: 查询出每年支付时间间隔最长的用户
select c.years,c.user_name,c.pay_days 
from(
select b.years,b.user_name,datediff(b.pay_time,b.lg) as pay_days,
rank() over(partition by b.years order by datediff(b.pay_time,b.lg) desc) as rk 
from (
select year(a.pay_time) as years,a.user_name,a.pay_time,
lag(a.pay_time) over(partition by a.user_name,year(a.pay_time) order by a.pay_time) as lg
from user_trade a 
) b 
where b.lg is not null
) c 
where c.rk = 1;