Oracle SQL从有序数据集中获取第一个和最后一个记录

Question

我正在处理的软件需要获取有序数据集的第一个和最后一个记录.数据集按日期列排序.

我有的数据:

--table "notes":
--    ordered by this
--                |
--                V
note_id      date_created attribute1  attribute2  ... -- I want to get
-----------------------------------------------------
596          2014/01/20   ...         ...         ... -- <- this
468          2014/02/28   ...         ...         ...
324          2014/03/01   ...         ...         ...
532          2014/04/08   ...         ...         ...
465          2014/05/31   ...         ...         ... -- <- and this

期望的输出:

596          2014/01/20   ...         ...         ...
465          2014/05/31   ...         ...         ...

Answer 1

您可以使用窗口功能:

select t.*
from (select t.*, row_number() over (order by date_created) as seqnum,
             count(*) over () as cnt
      from t
     ) t
where seqnum = 1 or seqnum = cnt;

在Oracle 12中,您还可以:

select t.*
from t
order by date_created
fetch first 1 rows only
union all
select t.*
from t
order by date_created desc
fetch first 1 rows only;

Answer 2

如果我猜对了，请尝试以下操作：

select t1.*
  from YOUR_TABLE t1
     , (
        select min(note_id) keep(dense_rank first order by date_created) min_val
             , max(note_id) keep(dense_rank last order by date_created) max_val
          from YOUR_TABLE
       ) t2
 where t1.note_id = t2.min_val
    or t1.note_id = t2.max_val