oracle分组后拼接

-----关于分组后字段拼接的问题

来自:www.itpub.net最近在论坛上,经常会看到关于分组后字段拼接的问题,
大概是类似下列的情形:

SQL> select no,q from test
2 /

NO Q
———- ——————————
001 n1
001 n2
001 n3
001 n4
001 n5
002 m1
003 t1
003 t2
003 t3
003 t4
003 t5
003 t6

12 rows selected

最后要得到类似于如下的结果:
001 n1;n2;n3;n4;n5
002 m1
003 t1;t2;t3;t4;t5;t6

通常大家都认为这类问题无法用一句SQL解决,本来我也这么认为,可是今天无意中突然有了灵感,原来是可以这么做的:
前几天有人提到过sys_connect_by_path的用法,我想这里是不是也能用到这个方法,如果能做到的话,不用函数或存贮过程也可以做到了;要用到sys_connect_by_path,首先要自己构建树型的结构,并且树的每个分支都是单根的,例如1-〉2-〉3-〉4,不会存在1-〉2,1-〉3的情况;
我是这么构建树,很简单的,看下面的结果就会知道了:
SQL> select no,q,rn,lead(rn) over(partition by no order by rn) rn1
2 from (select no,q,row_number() over(order by no,q desc) rn from test)
3 /

NO Q RN RN1
———- —————————— ———- ———-
001 n5 1 2
001 n4 2 3
001 n3 3 4
001 n2 4 5
001 n1 5
002 m1 6
003 t6 7 8
003 t5 8 9
003 t4 9 10
003 t3 10 11
003 t2 11 12
003 t1 12

12 rows selected

有了这个树型的结构,接下来的事就好办了,只要取出拥有全路径的那个path,问题就解决了,先看no=‘001’的分组:
select no,sys_connect_by_path(q,’;’) result from
(select no,q,rn,lead(rn) over(partition by no order by rn) rn1
from (select no,q,row_number() over(order by no,q desc) rn from test)
)
start with no = ‘001’ and rn1 is null connect by rn1 = prior rn
SQL>
6 /

NO RESULT
———- ——————————————————————————–
001 ;n1
001 ;n1;n2
001 ;n1;n2;n3
001 ;n1;n2;n3;n4
001 ;n1;n2;n3;n4;n5

上面结果的最后1条就是我们要得结果了
要得到每组的结果,可以下面这样

代码:


select t
.*,
(
select max(sys_connect_by_path(q,';')) result from
(select no,q,rn,lead(rn) over(partition by no order by rn) rn1
from
(select no,q,row_number() over(order by no,q desc) rn from test)
)
start with no = t.no and rn1 is null connect by rn1 = prior rn
) value
from
(select distinct no from test)  


 

SQL>
10 /

NO VALUE
———- ——————————————————————————–
001 ;n1;n2;n3;n4;n5
002 ;m1
003 ;t1;t2;t3;t4;t5;t6

对上面结果稍加处理就可以了,希望对大家有帮助:)

稍微改进下:

代码:

 


select no,max(sys_connect_by_path(q,';')) result from ( select no,q,rn,lead(rnover(partition by no order by rnrn1 from ( select no,q,row_number() over(order by no,q descrn from test ) ) start with rn1 is null connect by rn1 prior rn group by no



 

改进下算法,少一层嵌套查询,效率会好些:

 

 

代码:


select no
,max(sys_connect_by_path(q,';')) result from
(
select no,q,(row_number() over(order by no,q desc) + rank() over(order by no)) rn
from test
)
connect by rn-1 = prior rn
group by no;


 

 

 

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