今天兄弟们的帮助，数据库，具有数据如下面的表：
no name
1 a
2 b
3 c
4 d
怎样用一个sql显演示样例如以下结果：
ab
ac
ad
bc
bd
cd
对于这样的构造数据，是分析函数的强项。以下来做个试验：
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
实现3：也能够不用分析函数
select a.name || b.name from t a, t b where a.no < b.no;
附录，一句SQL实现9*9乘法口诀：
select r1 || ‘*‘ || r1 || ‘=‘ || r1 * r1 A,
decode(r2, ‘‘, ‘‘, r2 || ‘*‘ || r1 || ‘=‘ || r2 * r1) b,
decode(r3, ‘‘, ‘‘, r3 || ‘*‘ || r1 || ‘=‘ || r3 * r1) C,
decode(r4, ‘‘, ‘‘, r4 || ‘*‘ || r1 || ‘=‘ || r4 * r1) D,
decode(r5, ‘‘, ‘‘, r5 || ‘*‘ || r1 || ‘=‘ || r5 * r1) E,
decode(r6, ‘‘, ‘‘, r6 || ‘*‘ || r1 || ‘=‘ || r6 * r1) F,
decode(r7, ‘‘, ‘‘, r7 || ‘*‘ || r1 || ‘=‘ || r7 * r1) G,
decode(r8, ‘‘, ‘‘, r8 || ‘*‘ || r1 || ‘=‘ || r8 * r1) H,
decode(r9, ‘‘, ‘‘, r9 || ‘*‘ || r1 || ‘=‘ || r9 * r1) I
from (select level r1,
lag(level, 1) over(order by level) r2,
lag(level, 2) over(order by level) r3,
lag(level, 3) over(order by level) r4,
lag(level, 4) over(order by level) r5,
lag(level, 5) over(order by level) r6,
lag(level, 6) over(order by level) r7,
lag(level, 7) over(order by level) r8,
lag(level, 8) over(order by level) r9
from dual
connect by level < 10);