sql-server - 仅检索不同的组合

Question

我Node varchar(25)在 MS-SQL Server 中有列。可能的值为

node
-----
D-C
C-B
B-A
B-C
B-E
C-A 
A-B
A-C
C-D
etc.

我想从中检索不同的组合。例如：

node
----
D-C
C-B
B-A
B-E
C-A

请为此告诉SQL。

Answer 1

您将两条数据压入一列。这并不理想。所以我的解决方案首先必须纠正这个：

SQL> create table mytable (node)
  2  as
  3  select 'D-C' from dual union all
  4  select 'C-B' from dual union all
  5  select 'B-A' from dual union all
  6  select 'B-C' from dual union all
  7  select 'B-E' from dual union all
  8  select 'C-A' from dual union all
  9  select 'A-B' from dual union all
 10  select 'A-C' from dual union all
 11  select 'C-D' from dual
 12  /

Table created.

SQL> with structured_data as
  2  ( select regexp_substr(node,'[^-]+',1,1) startnode
  3         , regexp_substr(node,'[^-]+',1,2) endnode
  4      from mytable
  5  )
  6  select distinct least(startnode,endnode) startnode
  7       , greatest(startnode,endnode) endnode
  8    from structured_data
  9  /

STARTNODE ENDNODE
--------- -------
B         E
A         C
A         B
C         D
B         C

5 rows selected.

Answer 2

select distinct(Node) from YOUR_TABLE;

这是一个带有示例的 SQLfiddle：http ://www.sqlfiddle.com/#!4/76484/2

我回答了 oracle，因为这个问题被标记为 oracle。对于 MS-Server，它可能是同样的事情......

Answer 3

另一种方法。非常类似于 Rob van Wijk 在使用greatest()和least()函数方面发布的内容，但没有调用正则表达式函数。我们可以计算最大和最小列值及其反转值 - 函数返回的值reverse()：

注意： reverse()函数是未记录的函数。不要在生产应用程序中使用它。

with t1(col) as(
   select 'D-C' from dual union all
   select 'C-B' from dual union all
   select 'B-A' from dual union all
   select 'B-C' from dual union all
   select 'B-E' from dual union all
   select 'C-A' from dual union all
   select 'A-B' from dual union all
   select 'A-C' from dual union all
   select 'D-C' from dual
 )
 select res_1  /* as well/or we can choose res_2 */
   from (select distinct
                greatest(col, reverse(col)) as res_1
              , least(col, reverse(col))    as res_2
          from t1)

结果：

RES_1
-----
D-C
B-A
C-A
C-B
E-B

或#2

select col
  from ( select col
              , row_number() over(partition by greatest(col, reverse(col))
                                             , least(col, reverse(col)) 
                                      order by col) as rn
          from t1
       )
 where rn = 1

结果：

COL
---
A-B
A-C
B-C
D-C
B-E