日期:2014-05-16  浏览次数:20591 次

用lead分析函数构造类似9*9乘法口诀的功能
今天又个兄弟求助,数据库里一个表有数据如下:
no  name
1    a
2    b
3    c
4    d
如何用一个sql显示如下结果:
ab
ac
ad
bc
bd

cd


对于这种构造数据,是分析函数的强项,下面来做个试验:

create table t (no number,name varchar(2));
insert into t values(1,'a');
insert into t values(2,'b');
insert into t values(3,'c');
insert into t values(4,'d');
commit;

实现1:
select decode(h2, '', '', h1 || h2) b,
       decode(h3, '', '', h1 || h3) c,
       decode(h4, '', '', h1 || h4) d
  from (select name h1,
               lead(name, 1) over(order by name) h2,
               lead(name, 2) over(order by name) h3,
               lead(name, 3) over(order by name) h4
          from t) ;

B    C    D
---- ---- ----
ab   ac   ad
bc   bd
cd

实现2:相对实现1对于行进行了转换
with tt as(
select name h1,
               lead(name, 1) over(order by name) h2,
               lead(name, 2) over(order by name) h3,
               lead(name, 3) over(order by name) h4
          from t
)
select * from (select decode(h2, '', '', h1 || h2) b from tt
union
select decode(h3, '', '', h1 || h3) c from tt
union
select decode(h4, '', '', h1 || h4) d from tt)
where b is not null;

B
----
ab
ac
ad
bc
bd
cd