Oracle排名函数(Rank)实例详解

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.名次<=2

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)
);
insert into t_score (autoid, s_id, s_name, sub_name, score)
values (8, 1, ‘张三 ‘, ‘语文’, 80);
insert into t_score (autoid, s_id, s_name, sub_name, score)
values (9, 2, ‘李四 ‘, ‘数学’, 80);
insert into t_score (autoid, s_id, s_name, sub_name, score)
values (10, 1, ‘张三 ‘, ‘数学’, 0);
insert into t_score (autoid, s_id, s_name, sub_name, score)
values (11, 2, ‘李四 ‘, ‘语文’, 50);
insert into t_score (autoid, s_id, s_name, sub_name, score)
values (12, 3, ‘张三丰 ‘, ‘语文’, 10);
insert into t_score (autoid, s_id, s_name, sub_name, score)
values (13, 3, ‘张三丰 ‘, ‘数学’, null);
insert into t_score (autoid, s_id, s_name, sub_name, score)
values (14, 3, ‘张三丰 ‘, ‘体育’, 120);
insert into t_score (autoid, s_id, s_name, sub_name, score)
values (15, 4, ‘杨过 ‘, ‘java’, 90);
insert into t_score (autoid, s_id, s_name, sub_name, score)
values (16, 5, ‘mike ‘, ‘c++’, 80);
insert into t_score (autoid, s_id, s_name, sub_name, score)
values (3, 3, ‘张三丰 ‘, ‘oracle’, 0);
insert into t_score (autoid, s_id, s_name, sub_name, score)
values (4, 4, ‘杨过 ‘, ‘oracle’, 77);
insert into t_score (autoid, s_id, s_name, sub_name, score)
values (17, 2, ‘李四 ‘, ‘oracle’, 77);
commit;