日期:2014-05-17  浏览次数:20897 次

求一个行列转换的写法(ORACLE),多谢
求一个行列转换的写法(ORACLE)

create table t_testHL
(
number varchar2(20),
node varchar2(20),
qty number(6)
bizDate Date
);

有以下数据
bizDate number node qty
2009-12-1 A 工序A 30
2009-12-1 B 工序C 40
2009-12-2 A 工序B 30
2009-12-3 A 工序C 40
2009-12-4 B 工序D 30
2009-12-5 A 工序A 10

 
bizDate从2009-12-1到2009-12-4结果为
number 工序A 工序B 工序C 工序D  
  A 30 30 40
  B 40 30
 合计 30 30 80 30


bizDate从2009-12-1到2009-12-5结果为
number 工序A 工序B 工序C 工序D
 A 40 30 40
 B 40 30 
合计 40 30 80 30

bizDate从2009-12-1到2009-12-3结果为
number 工序A 工序B 工序C  
 A 30 30 40
 B 40  
合计 30 30 80  

如果横向合计也能出来的话 那更好,多谢了!!

------解决方案--------------------
SQL code
with t_dec as(
select to_date('2009-12-1','yyyy-mm-dd') bizDate, 'A' number_, '工序A' node, 30 qty from dual union all
select to_date('2009-12-1','yyyy-mm-dd') bizDate, 'B' number_, '工序C' node, 40 qty from dual union all
select to_date('2009-12-2','yyyy-mm-dd') bizDate, 'A' number_, '工序B' node, 30 qty from dual union all
select to_date('2009-12-3','yyyy-mm-dd') bizDate, 'A' number_, '工序C' node, 40 qty from dual union all
select to_date('2009-12-4','yyyy-mm-dd') bizDate, 'B' number_, '工序D' node, 30 qty from dual union all
select to_date('2009-12-5','yyyy-mm-dd') bizDate, 'A' number_, '工序A' node, 10 qty from dual)
select number_,
sum(decode(node,'工序A',qty)) 工序A,
sum(decode(node,'工序B',qty)) 工序B,
sum(decode(node,'工序C',qty)) 工序C,
sum(decode(node,'工序D',qty)) 工序D
from t_dec
where bizDate between to_date('2009-12-1','yyyy-mm-dd') and to_date('2009-12-5','yyyy-mm-dd')
group by number_
union all
select '合计' bizDate, sum(工序A), sum(工序B), sum(工序C), sum(工序D)
from(select number_,
sum(decode(node,'工序A',qty)) 工序A,
sum(decode(node,'工序B',qty)) 工序B,
sum(decode(node,'工序C',qty)) 工序C,
sum(decode(node,'工序D',qty)) 工序D
from t_dec
where bizDate between to_date('2009-12-1','yyyy-mm-dd') and to_date('2009-12-5','yyyy-mm-dd')
group by number_);