Q)从每个组中找到TOP X记录(Oracle)

SQL> select * from emp e 
  2  where e.empno in (select d.empno from emp d 
  3  where d.deptno=e.deptno and rownum<3)
  4  order by deptno
  5  ;

 EMPNO ENAME      JOB              MGR HIREDATE         SAL       COMM     DEPTNO

  7782 CLARK      MANAGER         7839 09-JUN-81       2450                    10
  7839 KING       PRESIDENT            17-NOV-81       5000                    10
  7369 SMITH      CLERK           7902 17-DEC-80        800                    20
  7566 JONES      MANAGER         7839 02-APR-81       2975                    20
  7499 ALLEN      SALESMAN        7698 20-FEB-81       1600        300         30
  7521 WARD       SALESMAN        7698 22-FEB-81       1250        500         30

选定6行。

UNION操作符对您有用吗?每个部分有一个SELECT，然后将它们联合在一起。不过，我猜它只适用于固定数量的部分。

这适用于SQL Server 2005(编辑以反映您的澄清):

select *
from Things t
where t.ThingID in (
    select top 10 ThingID
    from Things tt
    where tt.Section = t.Section and tt.ThingDate = @Date
    order by tt.DateEntered desc
    )
    and t.ThingDate = @Date
order by Section, DateEntered desc

如果你知道这些部分是什么，你可以这样做:

select top 10 * from table where section=1
union
select top 10 * from table where section=2
union
select top 10 * from table where section=3

如果你正在使用SQL 2005，你可以这样做…

SELECT rs.Field1,rs.Field2 
    FROM (
        SELECT Field1,Field2, Rank() 
          over (Partition BY Section
                ORDER BY RankCriteria DESC ) AS Rank
        FROM table
        ) rs WHERE Rank <= 10

如果你的RankCriteria有平局，那么你可能会返回超过10行，Matt的解决方案可能更适合你。

我知道这个帖子有点老了，但我刚刚遇到了一个类似的问题(从每个类别中选择最新的文章)，这是我想出的解决方案:

WITH [TopCategoryArticles] AS (
    SELECT 
        [ArticleID],
        ROW_NUMBER() OVER (
            PARTITION BY [ArticleCategoryID]
            ORDER BY [ArticleDate] DESC
        ) AS [Order]
    FROM [dbo].[Articles]
)
SELECT [Articles].* 
FROM 
    [TopCategoryArticles] LEFT JOIN 
    [dbo].[Articles] ON
        [TopCategoryArticles].[ArticleID] = [Articles].[ArticleID]
WHERE [TopCategoryArticles].[Order] = 1

这与Darrel的解决方案非常相似，但克服了可能返回比预期更多行的RANK问题。