Question

我在postgresql-9.1.x数据库中有一个表，其定义如下：

# \d cms
                                      Table "public.cms"
   Column    |            Type             |                    Modifiers                     
-------------+-----------------------------+--------------------------------------------------
 id          | integer                     | not null default nextval('cms_id_seq'::regclass)
 last_update | timestamp without time zone | not null default now()
 system      | text                        | not null
 owner       | text                        | not null
 action      | text                        | not null
 notes       | text

以下是表格中的数据样本：

 id  |        last_update         |        system        |   owner   |               action                | 
     notes
 ----+----------------------------+----------------------+-----------+-------------------------------------    +-
----------------
 584 | 2012-05-04 14:20:53.487282 | linux32-test5   | rfell     | replaced MoBo/CPU                   | 
  34 | 2011-03-21 17:37:44.301984 | linux-gputest13 | apeyrovan | System deployed with production GPU | 
 636 | 2012-05-23 12:51:39.313209 | mac64-cvs11     | kbhatt    | replaced HD                         | 
 211 | 2011-09-12 16:58:16.166901 | linux64-test12  | rfell     | HD swap                             | 
drive too small

我想做的是制作一个SQL查询，它只返回系统和所有者列中的唯一/不同值（如果一列结果中的值数小于另一列的结果，则填入NULL ），而忽略了它们之间的关联。所以像这样：

 system          |    owner
-----------------+------------------
 linux32-test5   |   apeyrovan
 linux-gputest13 |   kbhatt 
 linux64-test12  |   rfell
 mac64-cvs11     |

我可以找到获取此数据的唯一方法是使用两个单独的SQL查询： SELECT系统FROM cms GROUP BY系统; SELECT owner FROM cms GROUP BY所有者;

Answer 1

很难从我这里询问你为什么要做这样的事情。以下查询通过使用row_number（）函数在计算列上执行连接来执行此操作：

select ts.system, town.owner
from (select system, row_number() over (order by system) as seqnum
      from (select distinct system
            from t
           ) ts
     ) ts full outer join
     (select owner, row_number() over (order by owner) as seqnum
      from (select distinct owner
            from t
           ) town
     ) town
     on ts.seqnum = town.seqnum

完整的外部联接可确保完整返回两个列表中较长的一个。

从多个无关联列中获取唯一/不同的值

1 个答案: