Oracle排名函数(Rank)实例详解
这篇文章主要介绍了Oracle排名函数(Rank)实例详解,需要的朋友可以参考下
--已知：两种排名方式(分区和不分区):使用和不使用partition
--两种计算方式(连续,不连续),对应函数：dense_rank,rank
·查询原始数据：学号,姓名,科目名,成绩
select * from t_score
| S_ID | S_NAME | SUB_NAME | SCORE |
| 1 | 张三 | 语文 | 80.00 |
| 2 | 李四 | 数学 | 80.00 |
| 1 | 张三 | 数学 | 0.00 |
| 2 | 李四 | 语文 | 50.00 |
| 3 | 张三丰 | 语文 | 10.00 |
| 3 | 张三丰 | 数学 | |
| 3 | 张三丰 | 体育 | 120.00 |
| 4 | 杨过 | JAVA | 90.00 |
| 5 | mike | c++ | 80.00 |
| 3 | 张三丰 | Oracle | 0.00 |
| 4 | 杨过 | Oracle | 77.00 |
| 2 | 李四 | Oracle | 77.00 |
·查询各学生科目为Oracle排名(简单排名)
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 | 名次 |
| 4 | 杨过 | Oracle | 77.00 | 1 |
| 2 | 李四 | Oracle | 77.00 | 1 |
| 3 | 张三丰 | Oracle | 0.00 | 3 |
对比：rank()与dense_rank()：非连续排名与连续排名(都是简单排名)
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 | 名次 |
| 4 | 杨过 | Oracle | 77.00 | 1 |
| 2 | 李四 | Oracle | 77.00 | 1 |
| 3 | 张三丰 | Oracle | 0.00 | 2 |
·查询各学生各科排名(分区排名)
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 | 名次 |
| 4 | 杨过 | JAVA | 90.00 | 1 |
| 4 | 杨过 | Oracle | 77.00 | 1 |
| 2 | 李四 | Oracle | 77.00 | 1 |
| 3 | 张三丰 | Oracle | 0.00 | 3 |
| 5 | mike | c++ | 80.00 | 1 |
| 3 | 张三丰 | 数学 | 1 | |
| 2 | 李四 | 数学 | 80.00 | 2 |
| 1 | 张三 | 数学 | 0.00 | 3 |
| 3 | 张三丰 | 体育 | 120.00 | 1 |
| 1 | 张三 | 语文 | 80.00 | 1 |
| 2 | 李四 | 语文 | 50.00 | 2 |
| 3 | 张三丰 | 语文 | 10.00 | 3 |
·查询各科前2名(分区排名)
·类似：新闻表,求栏目点击率在前3位的新闻。
商品表,求各类别销售额在前10位的商品。
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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 |
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S_ID
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S_NAME
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SUB_NAME
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SCORE
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名次
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4
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杨过
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JAVA
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90.00
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1
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4
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杨过
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Oracle
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77.00
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1
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2
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李四
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Oracle
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77.00
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1
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3
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张三丰
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Oracle
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0.00
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2
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5
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mike
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c++
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80.00
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1
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3
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张三丰
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数学
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1
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2
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李四
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数学
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80.00
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2
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3
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张三丰
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体育
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120.00
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1
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1
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张三
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语文
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80.00
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1
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2
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李四
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语文
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50.00
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2
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·查询各同学总分
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 | 张三 | 80.00 |
| 2 | 李四 | 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：
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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; |
