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) xwhere 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;