mysql8学习笔记⑨窗口函数

前言

MySQL8.0之前，做数据排名统计等相当痛苦，因为没有像Oracle、SQL SERVER 、PostgreSQL等其他数据库那样的窗口函数。但随着MySQL8.0中新增了窗口函数之后，针对这类统计就再也不是事了，本文就以常用的排序实例介绍MySQL的窗口函数。

1、准备工作

创建表及测试数据

 

mysql> create database testdb;
Database changed
/* 创建表 */
 create table tb_score(id int primary key auto_increment,stu_no varchar(10),course varchar(50),score decimal(4,1),key idx_stuNo_course(stu_no,course));


mysql> show tables;
+------------------+
| Tables_in_testdb |
+------------------+
| tb_score  |
+------------------+

/* 新增一批测试数据 */
insert into tb_score(stu_no,course,score)values('2020001','mysql',90),('2020001','C++',85),('2020003','English',100),('2020002','mysql',50),('2020002','C++',70),('2020002','English',99);
insert into tb_score(stu_no,course,score)values('2020003','mysql',78),('2020003','C++',81),('2020003','English',80),('2020004','mysql',80),('2020004','C++',60),('2020004','English',100);
insert into tb_score(stu_no,course,score)values('2020005','mysql',98),('2020005','C++',96),('2020005','English',70),('2020006','mysql',60),('2020006','C++',90),('2020006','English',70);
insert into tb_score(stu_no,course,score)values('2020007','mysql',50),('2020007','C++',66),('2020007','English',76),('2020008','mysql',90),('2020008','C++',69),('2020008','English',86);
insert into tb_score(stu_no,course,score)values('2020009','mysql',70),('2020009','C++',66),('2020009','English',86),('2020010','mysql',75),('2020010','C++',76),('2020010','English',81);
insert into tb_score(stu_no,course,score)values('2020011','mysql',90),('2020012','C++',85),('2020011','English',84),('2020012','English',75),('2020013','C++',96),('2020013','English',88);

2、统计每门课程分数的排名

根据每门课程的分数从高到低进行排名，此时，会出现分数相同时怎么处理的问题，下面就根据不同的窗口函数来处理不同场景的需求

ROW_NUMBER

由结果可以看出，分数相同时按照学号顺序进行排名

mysql> select stu_no,course,score,row_number() over(PARTITION by course order by score desc) as rn from tb_score;
+---------+---------+-------+----+
| stu_no  | course  | score | rn |
+---------+---------+-------+----+
| 2020005 | C++     | 96.0  |  1 |
| 2020013 | C++     | 96.0  |  2 |
| 2020006 | C++     | 90.0  |  3 |
| 2020001 | C++     | 85.0  |  4 |
| 2020012 | C++     | 85.0  |  5 |
| 2020003 | C++     | 81.0  |  6 |
| 2020010 | C++     | 76.0  |  7 |
| 2020002 | C++     | 70.0  |  8 |
| 2020008 | C++     | 69.0  |  9 |
| 2020007 | C++     | 66.0  | 10 |
| 2020009 | C++     | 66.0  | 11 |
| 2020004 | C++     | 60.0  | 12 |
| 2020003 | English | 100.0 |  1 |
| 2020004 | English | 100.0 |  2 |
| 2020002 | English | 99.0  |  3 |
| 2020013 | English | 88.0  |  4 |
| 2020008 | English | 86.0  |  5 |
| 2020009 | English | 86.0  |  6 |
| 2020011 | English | 84.0  |  7 |
| 2020010 | English | 81.0  |  8 |
| 2020003 | English | 80.0  |  9 |
| 2020007 | English | 76.0  | 10 |
| 2020012 | English | 75.0  | 11 |
| 2020005 | English | 70.0  | 12 |
| 2020006 | English | 70.0  | 13 |
| 2020005 | mysql   | 98.0  |  1 |
| 2020001 | mysql   | 90.0  |  2 |
| 2020008 | mysql   | 90.0  |  3 |
| 2020011 | mysql   | 90.0  |  4 |
| 2020004 | mysql   | 80.0  |  5 |
| 2020003 | mysql   | 78.0  |  6 |
| 2020010 | mysql   | 75.0  |  7 |
| 2020009 | mysql   | 70.0  |  8 |
| 2020006 | mysql   | 60.0  |  9 |
| 2020002 | mysql   | 50.0  | 10 |
| 2020007 | mysql   | 50.0  | 11 |
+---------+---------+-------+----+

DENSE_RANK

为了让分数相同时排名也相同，则可以使用DENSE_RANK函数，结果如下：

mysql> select stu_no,course,score,DENSE_RANK() over(partition by course order by score desc) rn from tb_score;
+---------+---------+-------+----+
| stu_no  | course  | score | rn |
+---------+---------+-------+----+
| 2020005 | C++     | 96.0  |  1 |
| 2020013 | C++     | 96.0  |  1 |
| 2020006 | C++     | 90.0  |  2 |
| 2020001 | C++     | 85.0  |  3 |
| 2020012 | C++     | 85.0  |  3 |
| 2020003 | C++     | 81.0  |  4 |
| 2020010 | C++     | 76.0  |  5 |
| 2020002 | C++     | 70.0  |  6 |
| 2020008 | C++     | 69.0  |  7 |
| 2020007 | C++     | 66.0  |  8 |
| 2020009 | C++     | 66.0  |  8 |
| 2020004 | C++     | 60.0  |  9 |
| 2020003 | English | 100.0 |  1 |
| 2020004 | English | 100.0 |  1 |
| 2020002 | English | 99.0  |  2 |
| 2020013 | English | 88.0  |  3 |
| 2020008 | English | 86.0  |  4 |
| 2020009 | English | 86.0  |  4 |
| 2020011 | English | 84.0  |  5 |
| 2020010 | English | 81.0  |  6 |
| 2020003 | English | 80.0  |  7 |
| 2020007 | English | 76.0  |  8 |
| 2020012 | English | 75.0  |  9 |
| 2020005 | English | 70.0  | 10 |
| 2020006 | English | 70.0  | 10 |
| 2020005 | mysql   | 98.0  |  1 |
| 2020001 | mysql   | 90.0  |  2 |
| 2020008 | mysql   | 90.0  |  2 |
| 2020011 | mysql   | 90.0  |  2 |
| 2020004 | mysql   | 80.0  |  3 |
| 2020003 | mysql   | 78.0  |  4 |
| 2020010 | mysql   | 75.0  |  5 |
| 2020009 | mysql   | 70.0  |  6 |
| 2020006 | mysql   | 60.0  |  7 |
| 2020002 | mysql   | 50.0  |  8 |
| 2020007 | mysql   | 50.0  |  8 |
+---------+---------+-------+----+

RANK

DENSE_RANK的结果是分数相同时排名相同了，但是下一个名次是紧接着上一个名次的，如果2个并列的第1之后，下一个我想是第3名，则可以使用RANK函数实现

mysql> select stu_no,course,score,rank() over(partition by course order by score desc) rn from tb_score;
+---------+---------+-------+----+
| stu_no  | course  | score | rn |
+---------+---------+-------+----+
| 2020005 | C++     | 96.0  |  1 |
| 2020013 | C++     | 96.0  |  1 |
| 2020006 | C++     | 90.0  |  3 |
| 2020001 | C++     | 85.0  |  4 |
| 2020012 | C++     | 85.0  |  4 |
| 2020003 | C++     | 81.0  |  6 |
| 2020010 | C++     | 76.0  |  7 |
| 2020002 | C++     | 70.0  |  8 |
| 2020008 | C++     | 69.0  |  9 |
| 2020007 | C++     | 66.0  | 10 |
| 2020009 | C++     | 66.0  | 10 |
| 2020004 | C++     | 60.0  | 12 |
| 2020003 | English | 100.0 |  1 |
| 2020004 | English | 100.0 |  1 |
| 2020002 | English | 99.0  |  3 |
| 2020013 | English | 88.0  |  4 |
| 2020008 | English | 86.0  |  5 |
| 2020009 | English | 86.0  |  5 |
| 2020011 | English | 84.0  |  7 |
| 2020010 | English | 81.0  |  8 |
| 2020003 | English | 80.0  |  9 |
| 2020007 | English | 76.0  | 10 |
| 2020012 | English | 75.0  | 11 |
| 2020005 | English | 70.0  | 12 |
| 2020006 | English | 70.0  | 12 |
| 2020005 | mysql   | 98.0  |  1 |
| 2020001 | mysql   | 90.0  |  2 |
| 2020008 | mysql   | 90.0  |  2 |
| 2020011 | mysql   | 90.0  |  2 |
| 2020004 | mysql   | 80.0  |  5 |
| 2020003 | mysql   | 78.0  |  6 |
| 2020010 | mysql   | 75.0  |  7 |
| 2020009 | mysql   | 70.0  |  8 |
| 2020006 | mysql   | 60.0  |  9 |
| 2020002 | mysql   | 50.0  | 10 |
| 2020007 | mysql   | 50.0  | 10 |
+---------+---------+-------+----+

这样就实现了各种排序需求。

NTILE

NTILE函数的作用是对每个分组排名后，再将对应分组分成N个小组，例如

mysql> select stu_no,course,score,rank() over(partition by course order by score desc) rn,NTILE(2) over(partition by course order by score desc) rn_group from tb_score;
+---------+---------+-------+----+----------+
| stu_no  | course  | score | rn | rn_group |
+---------+---------+-------+----+----------+
| 2020005 | C++     | 96.0  |  1 |        1 |
| 2020013 | C++     | 96.0  |  1 |        1 |
| 2020006 | C++     | 90.0  |  3 |        1 |
| 2020001 | C++     | 85.0  |  4 |        1 |
| 2020012 | C++     | 85.0  |  4 |        1 |
| 2020003 | C++     | 81.0  |  6 |        1 |
| 2020010 | C++     | 76.0  |  7 |        2 |
| 2020002 | C++     | 70.0  |  8 |        2 |
| 2020008 | C++     | 69.0  |  9 |        2 |
| 2020007 | C++     | 66.0  | 10 |        2 |
| 2020009 | C++     | 66.0  | 10 |        2 |
| 2020004 | C++     | 60.0  | 12 |        2 |
| 2020003 | English | 100.0 |  1 |        1 |
| 2020004 | English | 100.0 |  1 |        1 |
| 2020002 | English | 99.0  |  3 |        1 |
| 2020013 | English | 88.0  |  4 |        1 |
| 2020008 | English | 86.0  |  5 |        1 |
| 2020009 | English | 86.0  |  5 |        1 |
| 2020011 | English | 84.0  |  7 |        1 |
| 2020010 | English | 81.0  |  8 |        2 |
| 2020003 | English | 80.0  |  9 |        2 |
| 2020007 | English | 76.0  | 10 |        2 |
| 2020012 | English | 75.0  | 11 |        2 |
| 2020005 | English | 70.0  | 12 |        2 |
| 2020006 | English | 70.0  | 12 |        2 |
| 2020005 | mysql   | 98.0  |  1 |        1 |
| 2020001 | mysql   | 90.0  |  2 |        1 |
| 2020008 | mysql   | 90.0  |  2 |        1 |
| 2020011 | mysql   | 90.0  |  2 |        1 |
| 2020004 | mysql   | 80.0  |  5 |        1 |
| 2020003 | mysql   | 78.0  |  6 |        1 |
| 2020010 | mysql   | 75.0  |  7 |        2 |
| 2020009 | mysql   | 70.0  |  8 |        2 |
| 2020006 | mysql   | 60.0  |  9 |        2 |
| 2020002 | mysql   | 50.0  | 10 |        2 |
| 2020007 | mysql   | 50.0  | 10 |        2 |
+---------+---------+-------+----+----------+

-- 窗口函数

-- row_number,rank,dense_rank之间的区别

with test(study_name,class_name,score) as(
select 'sqlercn','mysql',95
union all
select 'tom','mysql',99
union all
select 'jerry','mysql',99
union all
select 'gavin','mysql',98
union all
select 'sqlercn','postgresql',99
union all
select 'tom','postgresql',99
union all
select 'jerry','postgresql',98
)
select study_name,class_name,score
            ,row_number() over(partition by class_name order by score desc) as rw
            ,rank() over(partition by class_name order by score desc) as rk
            ,dense_rank() over(partition by class_name order by score desc) as drk
from test
order by class_name,rw;
排名显示的方式不同

-- 按学习人数对课程进行排名，并列出每类课程学习人数排名前3的课程名称，学习人数以及名次

with tmp as(
select class_name,title,score
        ,rank() over(partition by class_name order by score desc) as cnt
from imc_course a
join imc_class b on a.class_id = b.class_id
)
select * from tmp where cnt<=3;


-- 每门课程的学习人数占奔雷课程总学习人数的百分比

with tmp as(
select class_name,title,study_cnt
        ,sum(study_cnt) over(partition by class_name) as class_total
from imc_course a
join imc_class b on b.class_id = a.class_id
)
select class_name,title,concat(study_cnt/class_total*100,'%')
from tmp
order by class_name;

-- 学习人数等于1000人的课程有哪些，列出他们的课程标题和学习人数

select title,study_cnt
from imc_course
where study_cnt = 1000;

-- 学习人数大于1000人的课程有哪些，列出他们的课程标题和学习人数

select title,study_cnt
from imc_course
where study_cnt > 1000;

开发sql容易发生的问题

-- 查询出分类ID为5的课程名称和分类名称

错误一：
在on中使用and进行过滤
select a.title,b.class_name
from imc_course a
join imc_class b on a.class_id = b.class_id and a.class_id=5;

把内连接变更为左外连接，起不到过滤出我们需要的数据的效果
select a.title,b.class_name
from imc_course a 
left join imc_class b on a.class_id = b.class_id and a.class_id=5

使用where条件语句就没有这样的问题
select a.title,b.class_name
from imc_course a 
left join imc_class b on a.class_id = b.class_id
where b.class_id=5;

select *
from imc_course
where title in (select title from imc_class);

 

如何避免

这样就实现了各种排序需求。

NTILE

NTILE函数的作用是对每个分组排名后，再将对应分组分成N个小组，例如