产品部门有一个分析需求，简化后是个组合问题，简单表述如下：

表结构

c4列表示状态，c1，c2，c3只要一个不为空，c4就为Y

现在想知道每个组合的情况

比如为什么组合(c1、c2)的值是2，因为只有2行(第1行、第3行)数据满足c1、c2都不为空

相关测试sql代码如下：

SQL> create table t (

2 id varchar2(1),

3 c1 int,

4 c2 int,

5 c3 int,

6 c4 varchar2(1)

7 );

Table created.

SQL> insert into t values ('a',1,1,null,'Y');

1 row created.

SQL> insert into t values ('b',null,1,1,'Y');

1 row created.

SQL> insert into t values ('c',1,1,1,'Y');

1 row created.

SQL> insert into t values ('d',1,null,null,'Y');

1 row created.

SQL> commit;

Commit complete.

SQL> select replace(path, '.*') path, count(1) cnt

2 from (select *

3 from (select (sys_connect_by_path(pivot_char, '.*')) Path

4 from (select regexp_substr(csv.csvdata, '[^,]+', 1, level) Pivot_char

5 from (select 'C1,C2,C3' as csvdata from dual) csv

6 connect by level <= 3)

7 connect by prior Pivot_char < Pivot_char

8 order by length(path), path) a,

9 (select nvl2(c1, 'C1', null) || nvl2(c2, 'C2', null) ||

10 nvl2(c3, 'C3', null) col_not_null

11 from t) b

12 where regexp_instr(b.col_not_null, a.path) > 0)

13 group by path

14 /

PATH CNT

---------- ----------

C1 3

C2 3

C1C2 2

C2C3 2

C3 2

C1C3 1

C1C2C3 1

7 rows selected.

主要思路是构造所有可能的组合和表中实际数据关联匹配得出结果：

第一步：构造3行数据

SQL> select regexp_substr(csv.csvdata, '[^,]+', 1, level) Pivot_char

2 from (select 'C1,C2,C3' as csvdata from dual) csv

3 connect by level <= 3

4 /

PIVOT_CHAR

----------------

C1

C2

C3

第二步：3行数据所有组合情况，(.*)是为了后期正则匹配

SQL> select (sys_connect_by_path(pivot_char, '.*')) Path

2 from (select regexp_substr(csv.csvdata, '[^,]+', 1, level) Pivot_char

3 from (select 'C1,C2,C3' as csvdata from dual) csv

4 connect by level <= 3)

5 connect by prior Pivot_char < Pivot_char

6 order by length(path), path

7 /

PATH

--------------------------------------------------------------------------------

.*C1

.*C2

.*C3

.*C1.*C2

.*C1.*C3

.*C2.*C3

.*C1.*C2.*C3

第三步：表中实际数据的组合情况

SQL> select nvl2(c1, 'C1', null) || nvl2(c2, 'C2', null) || nvl2(c3, 'C3', null) col_not_null

2 from t

3 /

COL_NOT_NULL

------

C1C2

C2C3

C1C2C3

C1

第四步：可能的7种组合和表中实际数据的组合情况关联匹配出结果，如(C1C2)可以匹配(C1)(C2)(C1C2)三种情况，最后结果如下

SQL> select replace(path, '.*') path, count(1) cnt

2 from (select *

3 from (select (sys_connect_by_path(pivot_char, '.*')) Path

4 from (select regexp_substr(csv.csvdata, '[^,]+', 1, level) Pivot_char

5 from (select 'C1,C2,C3' as csvdata from dual) csv

6 connect by level <= 3)

7 connect by prior Pivot_char < Pivot_char

8 order by length(path), path) a,

9 (select nvl2(c1, 'C1', null) || nvl2(c2, 'C2', null) ||

10 nvl2(c3, 'C3', null) col_not_null

11 from t) b

12 where regexp_instr(b.col_not_null, a.path) > 0)

13 group by path

14 /

PATH CNT

---------- ----------

C1 3

C2 3

C1C2 2

C2C3 2

C3 2

C1C3 1

C1C2C3 1

7 rows selected.

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