我是这样做的:

SELECT a.* FROM articles AS a
  LEFT JOIN articles AS a2 
    ON a.section = a2.section AND a.article_date <= a2.article_date
GROUP BY a.article_id
HAVING COUNT(*) <= 10;

更新:这个GROUP BY的例子只适用于MySQL和SQLite，因为这些数据库在GROUP BY方面比标准SQL更允许。大多数SQL实现要求选择列表中不属于聚合表达式的所有列也在GROUP BY中。

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。

您可以尝试这种方法。 这个查询为每个国家返回10个人口最多的城市。

   SELECT city, country, population
   FROM
   (SELECT city, country, population, 
   @country_rank := IF(@current_country = country, @country_rank + 1, 1) AS country_rank,
   @current_country := country 
   FROM cities
   ORDER BY country, population DESC
   ) ranked
   WHERE country_rank <= 10;

SELECT r.*
FROM
(
    SELECT
        r.*,
        ROW_NUMBER() OVER(PARTITION BY r.[SectionID]
                          ORDER BY r.[DateEntered] DESC) rn
    FROM [Records] r
) r
WHERE r.rn <= 10
ORDER BY r.[DateEntered] DESC

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