只需从INNER JOIN返回一行

Question

我想仅从内连接返回第一行.我有两张桌子:

TABLE_X |  TABLE_Y
id      |  id   creationdate  xid 
1       |  1    01/01/2011    1
2       |  2    01/01/2011    1
3       |  3    31/12/2010    2
4       |  4    28/12/2010    3

表Y中的行可以具有相同的创建日期,因此我首先获得MAX(creationdate),然后从该集合获得MAX(id),例如:

SELECT  a.id,
        c.id,
        d.id,
        e.id,
        d.CREATIONDATE,
        a.REFNUMBER,
        a.DATECREATED,
        a.DESCRIPTION,
        e.CATEGORYCODE,
        e.OUTSTANDINGAM_MONAMT,
        e.PREVPAIDAMOUN_MONAMT,
        e.TOTALINCURRED_MONAMT,
        e.LOSSFROMDATE,
FROM 
TABLE_A a
INNER JOIN TABLE_B b ON (b.id = a.id)
INNER JOIN TABLE_C c ON (c.id = b.id)
INNER JOIN TABLE_D d ON
(
   c.i =
   (
      select
      d.id
      FROM TABLE_D
      WHERE TABLE_D.id = c.id
      AND TABLE_D.id =
      (
         select
         max(id)
         from TABLE_D t1
         where c_id = c.id
         and CREATIONDATE =
         (
            select
            max(CREATIONDATE)
            from TABLE_D t2
            where t2.c_id = t1.c_id
         )
      )
   ) 
)

INNER JOIN TABLE_E e ON
(
   d.i =
   (
      select
      e.d_id
      from TABLE_E
      where d_id = d.id
      AND id =
      (
         select
         max(id)
         from e t1
         where e.d_id = d.id
         and CREATIONDATE =
         (
            select
            max(CREATIONDATE)
            from TABLE_E t2
            where t2.d_id = t1.d_id
         )
      )
   )
)

当我自己调用它时,这可以工作,但是当我将它添加到INNER JOIN时,我为表Y中的每个匹配行获取一行.

我想要的是creationdate和id的最新记录,其中来自TABLE_X的xid = id.

Answer 1

“当我将它添加到内部联接时”？什么内连接？用什么内连接？这个问题严重不明确，但我认为你需要这个（我只使用视图来保持清晰，你可以轻松地将它们放在大括号中并构建一个大查询）：

-- latest pairs of (id,creation) per xid
create view latest_id_per_xid as
   select xid,max(id) id,max(creation) creation from table_y group by xid;

-- this view leaves only the rows from table_y that have the same (id,creation,idx)
-- as the newest rows identified in the former view (it's basically a semijoin)
-- you could also join on id alone 
create view latest_whole_y as
   select table_y.* from table_y natural join latest_id_per_xid;

-- now the answer is easy:
select * from table_x join latest_whole_y

我手头没有数据库来检查小错误，但它应该运行良好。（警告：最大的假设是您永远不会有具有新 ID 和旧日期的记录）