其语法为: RANK ( expr [, expr]... ) WITHIN GROUP ( ORDER BY expr [ DESC | ASC ] [NULLS { FIRST | LAST }] [, expr [ DESC | ASC ] [NULLS { FIRST | LAST }]]...参考栗子三。
表xgj:--建表create table XGJ(col1 NUMBER,col2 NUMBER)--插入数据insert into xgj (COL1, COL2)values (1, 1);insert into xgj (COL1, COL2)values (2, 1);insert into xgj (COL1, COL2)values (3, 2);insert into xgj (COL1, COL2)values (3, 1);insert into xgj (COL1, COL2)values (4, 1);insert into xgj (COL1, COL2)values (4, 2);insert into xgj (COL1, COL2)values (5, 2);insert into xgj (COL1, COL2)values (5, 2);insert into xgj (COL1, COL2)values (6, 2);--提交数据commit;数据规格如下:分析功能:列出Col2分组后根据Col1排序,并生成数字列. 比较实用于在成绩表中查出各科前几名的信息。SELECT a.*,RANK() OVER(PARTITION BY col2 ORDER BY col1) "Rank" FROM xgj a;
栗子2
TABLE:xgj_2 (分数 ,科目) create table xgj_2(score NUMBER,subject VARCHAR2(20))insert into xgj_2(SCORE, SUBJECT)values (80, "数学");insert into xgj_2(SCORE, SUBJECT)values (70, "语文");insert into xgj_2(SCORE, SUBJECT)values (90, "数学");insert into xgj_2(SCORE, SUBJECT)values (60, "数学");insert into xgj_2(SCORE, SUBJECT)values (100, "数学");insert into xgj_2(SCORE, SUBJECT)values (88, "语文");insert into xgj_2(SCORE, SUBJECT)values (65, "语文");insert into xgj_2(SCORE, SUBJECT)values (77, "语文");commit;数据规格如下: 现在我想要的结果是:(即想要每门科目的前3名的分数) 数学,100 数学,90 数学,80 语文,88 语文,77 语文,70SQL:select *from (select rank() over(partition by a.subject order by a.score desc) rk, a.*from xgj_2 a) t where t.rk <= 3;
栗子3
拿第二个栗子的表 作为数据源。合计功能:计算出数值(65,’语文’)在Orade By score ,subject排序下的排序值,也就是score=65,subject=语文在排序以后的位置SELECT RANK(65,"语文") WITHIN GROUP (ORDER BY score,subject ) "Rank" FROMxgj_2; 结果如下:
表xgj_3 数据规格: A B C a liu wang a jin shu a cai kai b yang du b lin ying b yao cai b yang 99 例如:当rank时为: select m.a,m.b,m.c,rank() over(partition by a order by b) RK from xgj_3 m A B C RK a cai kai 1 a jin shu 2 a liu wang 3 b lin ying 1 b yang du 2 b yang 99 2 b yao cai 4 而如果用dense_rank时为: select m.a,m.b,m.c,dense_rank() over(partition by a order by b) RK from xgj_3 m A B C RK a cai kai 1 a jin shu 2 a liu wang 3 b lin ying 1 b yang du 2 b yang 99 2更多Oracle相关信息见Oracle 专题页面 http://www.linuxidc.com/topicnews.aspx?tid=12本文永久更新链接地址