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 /
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
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) t
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:)
举例有表如下:货品ID 价格类型 价格 1 零售价 3.00 1 进货价 2.00 2 零售价 5.00 2 进货价 4.00 用什么方法能得到如下结果货品ID 零售价 进货价 1 3.00 2.00 2 5.00 4.00 select t1.货品ID,t1.零售价,t2.进货价 from (select 货品ID,价格 as 零售价 from table where 价格类型 = '零售价') t1, (select 货品ID,价格 as 进货价 from table where 价格类型 = '进货价') t2 where t1.货品ID = t2.货品ID
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
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) t
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:)
1 零售价 3.00
1 进货价 2.00
2 零售价 5.00
2 进货价 4.00
用什么方法能得到如下结果货品ID 零售价 进货价
1 3.00 2.00
2 5.00 4.00
select t1.货品ID,t1.零售价,t2.进货价 from
(select 货品ID,价格 as 零售价 from table where 价格类型 = '零售价') t1,
(select 货品ID,价格 as 进货价 from table where 价格类型 = '进货价') t2
where t1.货品ID = t2.货品ID
晕~