分析函数 over用法 之row_number() runk_number

分析函数用于计算基于组的某种聚合值,每个组返回多个行,而聚合函数每个组只返回一个行

表:

create table TB_SCORE

(
id NUMBER(10),
class VARCHAR2(10),
sname VARCHAR2(30),
score NUMBER(10)
)

插入数据代码如下

insert into tb_score (ID, CLASS, SNAME, SCORE)
values (1, '数学', '学生1', 22);

insert into tb_score (ID, CLASS, SNAME, SCORE)
values (2, '数学', '学生1', 66);

insert into tb_score (ID, CLASS, SNAME, SCORE)
values (3, '英语', '学生1', 88);

insert into tb_score (ID, CLASS, SNAME, SCORE)
values (4, '英语', '学生1', 77);

insert into tb_score (ID, CLASS, SNAME, SCORE)
values (5, '英语', '学生1', 88);

insert into tb_score (ID, CLASS, SNAME, SCORE)
values (6, '数学', '学生1', 7);

insert into tb_score (ID, CLASS, SNAME, SCORE)
values (7, '数学', '学生1', 88);

执行:SELECT * FROM tb_score,结果如下

1.row_number()

row_number() over (partition by colume1 order by colume2 desc) as columename

注释:

可以用来分组排序:

这个主要是可用来获取rowid,普通的获取rowid 时,如果条件上的rowid不是从1开始取值,就会报错,因为rowid取值的时候每一次都是从一开始计数,所以会用分析函数获取,在这种情况下用分析函数,里面的colume1 必须是唯一的,不然会重复.

eg:

(1)按科目分组,再按成绩倒序

SELECT t.*,  row_number()over (partition by t.class order by t.score desc) as 名次 FROM  tb_score t

结果如下:

(2)取每门课前两名的成绩:

SELECT * FROM
(
SELECT t.*,
row_number() over(partition by t.class order by t.score desc) as 名次
FROM tb_score t
) t1
where t1.名次<3

结果如下:

(3)取每门课成绩前两名的平均值

SELECT avg(t1.score) ,t1.class FROM
(
SELECT t.class,t.score,
row_number() over(partition by t.class order by t.score desc) as 名次
FROM tb_score t
) t1
where t1.名次<3
group by t1.class

结果如下:

2:ranK(),dense_rank();

(1) rank

SELECT t.class,t.score,
rank() over(partition by t.class order by t.score desc) as 名次
FROM tb_score t

结果如下

(2)dense_rank

SELECT t.class,t.score,
rank() over(partition by t.class order by t.score desc) as 名次
FROM tb_score t

结果如下:

总结:

row_number() over 排序时,没有并列,rank() over和dense_rank() over 有并列

但是,rank() over是并列后跳过,而dense_rank()over是并列后连续