解决方案：
create table t_20120512_a ( id varchar2(6) primary key, cost number(3), p2 number(3), p3 number(3) ) / create table t_20120512_b ( id varchar2(6) primary key, name varchar2(50), b varchar2(100) ) / insert into t_20120512_a values ('100',1,1,1); insert into t_20120512_a values ('101001',2,2,2); insert into t_20120512_a values ('101002',3,3,3); commit; insert into t_20120512_b values ('100','语文',null); insert into t_20120512_b values ('101','物理',null); insert into t_20120512_b values ('101001','电学',null); insert into t_20120512_b values ('101002','力学',null); commit; select a.id, b.name ,a.total from ( select distinct case when GROUPING(substr(id,1,3))=0 then substr(id,1,3) else id end id,sum(cost) total from t_20120512_a group by grouping SETS (id,substr(id,1,3)) ) a, t_20120512_b b where a.id=b.id;posted on 2012-05-12 22:51 蓝红石 阅读( ...) 评论( ...) 编辑 收藏
转载于:https://www.cnblogs.com/liuweihua/archive/2012/05/12/2497643.html
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