为每个类别选择前10个记录

联盟操作符适合你吗?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，因为这些数据库对于GROUP BY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。

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。

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。

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

在T-SQL中，我会这样做:

WITH TOPTEN AS (
SELECT *, ROW_NUMBER()
over (
PARTITION BY [group_by_field]
order by [prioritise_field]
) AS RowNo
FROM [table_name]
)
SELECT * FROM TOPTEN WHERE RowNo <= 10

如果你想生成按节分组的输出，只显示每个节中最前面的n记录，如下所示:

SECTION     SUBSECTION


deer        American Elk/Wapiti
deer        Chinese Water Deer
dog         Cocker Spaniel
dog         German Shephard
horse       Appaloosa
horse       Morgan

.．.那么下面的代码应该适用于所有SQL数据库。如果您想要前10，只需在查询的末尾将2更改为10。

select
x1.section
, x1.subsection
from example x1
where
(
select count(*)
from example x2
where x2.section = x1.section
and x2.subsection <= x1.subsection
) <= 2
order by section, subsection;

设置:

create table example ( id int, section varchar(25), subsection varchar(25) );


insert into example select 0, 'dog', 'Labrador Retriever';
insert into example select 1, 'deer', 'Whitetail';
insert into example select 2, 'horse', 'Morgan';
insert into example select 3, 'horse', 'Tarpan';
insert into example select 4, 'deer', 'Row';
insert into example select 5, 'horse', 'Appaloosa';
insert into example select 6, 'dog', 'German Shephard';
insert into example select 7, 'horse', 'Thoroughbred';
insert into example select 8, 'dog', 'Mutt';
insert into example select 9, 'horse', 'Welara Pony';
insert into example select 10, 'dog', 'Cocker Spaniel';
insert into example select 11, 'deer', 'American Elk/Wapiti';
insert into example select 12, 'horse', 'Shetland Pony';
insert into example select 13, 'deer', 'Chinese Water Deer';
insert into example select 14, 'deer', 'Fallow';

如果我们使用SQL Server >= 2005，那么我们只能用一个选择来解决这个任务:

declare @t table (
Id      int ,
Section int,
Moment  date
);


insert into @t values
(   1   ,   1   , '2014-01-01'),
(   2   ,   1   , '2014-01-02'),
(   3   ,   1   , '2014-01-03'),
(   4   ,   1   , '2014-01-04'),
(   5   ,   1   , '2014-01-05'),


(   6   ,   2   , '2014-02-06'),
(   7   ,   2   , '2014-02-07'),
(   8   ,   2   , '2014-02-08'),
(   9   ,   2   , '2014-02-09'),
(   10  ,   2   , '2014-02-10'),


(   11  ,   3   , '2014-03-11'),
(   12  ,   3   , '2014-03-12'),
(   13  ,   3   , '2014-03-13'),
(   14  ,   3   , '2014-03-14'),
(   15  ,   3   , '2014-03-15');




-- TWO earliest records in each Section


select top 1 with ties
Id, Section, Moment
from
@t
order by
case
when row_number() over(partition by Section order by Moment) <= 2
then 0
else 1
end;




-- THREE earliest records in each Section


select top 1 with ties
Id, Section, Moment
from
@t
order by
case
when row_number() over(partition by Section order by Moment) <= 3
then 0
else 1
end;




-- three LATEST records in each Section


select top 1 with ties
Id, Section, Moment
from
@t
order by
case
when row_number() over(partition by Section order by Moment desc) <= 3
then 0
else 1
end;

   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;

尝试了下面的方法，它也适用于领带。

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

虽然这个问题是关于SQL Server 2005的，但大多数人已经继续前进了，如果他们确实发现了这个问题，在其他情况下的首选答案可能是如本文所述，使用CROSS APPLY。

SELECT *
FROM t
CROSS APPLY (
SELECT TOP 10 u.*
FROM u
WHERE u.t_id = t.t_id
ORDER BY u.something DESC
) u

该查询涉及两个表。OP的查询只涉及1个表，在这种情况下，基于窗口函数的解决方案可能更有效。