MySQL实战窗口函数SQL分析班级学生考试成绩及生活消费

create table exam_score(
    sname varchar(20),
    age int,
    subject varchar(20),
    score varchar(20)
)charset=utf8;

# ----------------------- #

create table cost_fee(
    sname varchar(20),
    buydate varchar(20),
    buycost int
)charset=utf8;

insert into exam_score values
('张三' , 18, '语文' , 90),
('张三' , 18, '数学' , 80),
('张三' , 18, '英语' , 70),
('李四' , 21, '语文' , 88),
('李四' , 21, '数学' , 78),
('李四' , 21, '英语' , 71),
('王五' , 18, '语文' , 95),
('王五' , 18, '数学' , 83),
('王五' , 18, '英语' , 71),
('赵六' , 19, '语文' , 98),
('赵六' , 19, '数学' , 90),
('赵六' , 19, '英语' , 80);
# ----------------------- #
insert into cost_fee values
('张三','2019-01-01',10),
('张三','2019-03-03',23),
('张三','2019-02-05',46),
('李四','2019-02-02',15),
('李四','2019-01-07',50),
('李四','2019-03-04',29),
('王五','2019-03-08',62),
('王五','2019-02-09',68),
('王五','2019-01-11',75),
('赵六','2019-02-08',55),
('赵六','2019-03-10',12),
('赵六','2019-01-12',80);

select 
	sname
    ,subject
    ,score
    ,avg(score) over(partition by sname) as avg_score
from
	exam_score

select
	sname
    ,buydate
    ,buycost
    ,sum(buycost) over(partition by sname) as sum_cost
from
	cost_fee

select
	sname
    ,buydate
    ,buycost
    ,sum(buycost) over(partition by sname order by buydate) as sum_cost
from
	cost_fee

select
	sname
	,subject
	,score
    ,row_number() over(partition by subject order by score) rank1
from
	exam_score

select
	sname
	,subject
	,score
    ,rank() over(partition by subject order by score) rank1
from
	exam_score

select
	sname
	,subject
	,score
    ,dense_rank() over(partition by subject order by score) rank1
from
	exam_score

select
	sname
	,subject
	,score
    ,ntile(4) over() rank1
from
	exam_score

select
	sname
	,subject
    ,score
    ,ntile(4) over(partition by subject) rank1
from
	exam_score

select
	sname
	,subject
    ,score
    ,ntile(4) over(partition by subject order by score) rank1
from
	exam_score

select
	sname
	,buydate
    ,lag(buydate,1,'first day') over(partition by sname order by buydate) as 上次购买时间
    ,lead(buydate,1,'last day') over(partition by sname order by buydate) as 下次购买时间
from
	cost_fee

select
	sname
	,buydate
    ,first_value(buydate) over(partition by sname order by buydate) as 首次购买时间
    ,last_value(buydate) over(partition by sname order by buydate) as 最后一次购买时间
from
	cost_fee

select
	sname
	,buydate
    ,first_value(buydate) over(partition by sname order by buydate) as 首次购买时间
    ,last_value(buydate) over(partition by sname ) as 最后一次购买时间
from
	cost_fee