rank oracle window function of the ranking function

- known: two kinds are ranked ( partition and not partition): with and without partition

- two calculations ( continuous, discontinuous), the corresponding function: dense_rank, rank

· Raw data query: Student ID, name, account name, score

select * from t_score

S_ID	S_NAME	SUB_NAME	SCORE
1	Joe Smith	Chinese	80.00
2	John Doe	mathematics	80.00
1	Joe Smith	mathematics	0.00
2	John Doe	Chinese	50.00
3	Chi Master	Chinese	10.00
3	Chi Master	mathematics
3	Chi Master	physical education	120.00
4	Yang Guo	Java	90.00
5	mike	c++	80.00
3	Chi Master	Oracle	0.00
4	Yang Guo	oracle	77.00
2	John Doe	Oracle	77.00

· Oracle query subjects for each student ranking (simple ranking)

select sc.s_id,sc.s_name,sub_name,sc.score,

rank() over (order by score desc) 名次

from t_score sc

where sub_name='Oracle'

S_ID	S_NAME	SUB_NAME	SCORE	Ranking
4	Yang Guo	Oracle	77.00	1
2	John Doe	Oracle	77.00	1
3	Chi Master	Oracle	0.00	3

Contrast: rank () and dense_rank (): discontinuous and continuous Rank Rank (it is simple ranking)

select sc.s_id,sc.s_name,sub_name,sc.score,

dense_rank() over (order by score desc) 名次

from t_score sc

where sub_name='Oracle'

S_ID	S_NAME	SUB_NAME	SCORE	Ranking
4	Yang Guo	Oracle	77.00	1
2	John Doe	Oracle	77.00	1
3	Chi Master	Oracle	0.00	2

• The query subjects each student ranking ( partition rank)

select sc.s_id,sc.s_name,sub_name,sc.score,

rank() over

(partition by sub_name order by score desc) 名次

from t_score sc

S_ID	S_NAME	SUB_NAME	SCORE	Ranking
4	Yang Guo	JAVA	90.00	1
4	Yang Guo	Oracle	77.00	1
2	John Doe	Oracle	77.00	1
3	Chi Master	Oracle	0.00	3
5	mike	c++	80.00	1
3	Chi Master	mathematics		1
2	John Doe	mathematics	80.00	2
1	Joe Smith	mathematics	0.00	3
3	Chi Master	physical education	120.00	1
1	Joe Smith	Chinese	80.00	1
2	John Doe	Chinese	50.00	2
3	Chi Master	Chinese	10.00	3

• The query subjects before 2 name (for partition rank)

· Similar: News table, find the column hits the previous three news.

Product table, find the top 10 of each category of merchandise sales.

select * from (

select sc.s_id,sc.s_name,sub_name,sc.score,

dense_rank() over

(partition by sub_name order by score desc) 名次

from t_score sc

) x

where x. Place <= 2

S_ID	S_NAME	SUB_NAME	SCORE	Ranking
4	Yang Guo	JAVA	90.00	1
4	Yang Guo	Oracle	77.00	1
2	John Doe	Oracle	77.00	1
3	Chi Master	Oracle	0.00	2
5	mike	c++	80.00	1
3	Chi Master	mathematics		1
2	John Doe	mathematics	80.00	2
3	Chi Master	physical education	120.00	1
1	Joe Smith	Chinese	80.00	1
2	John Doe	Chinese	50.00	2

• The query each student scores

select s_id,s_name,sum(score) sum_score from t_score

group by s_id,s_name

S_ID	S_NAME	SUM_SCORE
1	Joe Smith	80.00
2	John Doe	207.00
3	张三丰	130.00
4	杨过	167.00
5	mike	80.00

·根据总分查询各同学名次

select x.*,

rank() over (order by sum_score desc) 名次

from (

select s_id,s_name,sum(score) sum_score from t_score

group by s_id,s_name ) x

S_ID	S_NAME	SUM_SCORE	名次
2	李四	207.00	1
4	杨过	167.00	2
3	张三丰	130.00	3
1	张三	80.00	4
5	mike	80.00	4

语法：

rank() over (order by 排序字段顺序)

rank() over (partition by 分组字段 order by 排序字段顺序)

1.顺序：asc|desc 名次与业务相关：

示例：找求优秀学员：成绩：降序迟到次数：升序

2.分区字段：根据什么字段进行分区。

问题：分区与分组有什么区别?

·分区只是将原始数据进行名次排列(记录数不变),

·分组是对原始数据进行聚合统计(记录数变少,每组返回一条),注意：聚合。

脚本：

create table t_score

(

autoid number primary key,

s_id number(3),

s_name char(8) not null,

sub_name varchar2(20),

score number(10,2)

);