각 카테고리마다 상위 10 개 레코드를 선택하십시오.

각 카테고리마다 상위 10 개 레코드를 선택하십시오.

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하나의 쿼리로 각 섹션에서 상위 10 개의 레코드를 반환하고 싶습니다. 누구나 그것을하는 방법을 도울 수 있습니까? 섹션은 테이블의 열 중 하나입니다.

데이터베이스는 SQL Server 2005입니다. 입력 한 날짜별로 상위 10 개를 반환하려고합니다. 섹션은 비즈니스, 지역 및 기능입니다. 특정 날짜의 경우 상위 10 개 비즈니스 행 (가장 최근 항목), 상위 10 개 로컬 행 및 상위 10 개 기능 만 원합니다.

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SQL 2005를 사용하는 경우 다음과 같이 할 수 있습니다 ...

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

RankCriteria에 관계가 있으면 10 개가 넘는 행을 반환 할 수 있으며 Matt의 솔루션이 더 나을 수 있습니다.

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T-SQL에서는 다음을 수행합니다.

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

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이것은 SQL Server 2005에서 작동합니다 (설명을 반영하도록 편집 됨).

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

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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

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나는 이것을 이렇게한다 :

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;

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업데이트 : 이 GROUP BY 예제는 MySQL 및 SQLite에서만 작동합니다. 데이터베이스는 GROUP BY에 관한 표준 SQL보다 더 관대하기 때문입니다. 대부분의 SQL 구현에서는 집계 표현식의 일부가 아닌 선택 목록의 모든 열도 GROUP BY에 있어야합니다.

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SQL Server> = 2005를 사용하면 한 번의 선택 만으로 작업을 해결할 수 있습니다 .

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;

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섹션이 무엇인지 아는 경우 다음을 수행 할 수 있습니다.

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

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나는이 스레드가 조금 오래되었다는 것을 알고 있지만 비슷한 문제에 봉착했습니다 (각 범주에서 최신 기사를 선택하십시오). 이것은 내가 생각해 낸 해결책입니다.

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

이것은 Darrel의 솔루션과 매우 유사하지만 의도 한 것보다 많은 행을 리턴 할 수있는 RANK 문제를 극복합니다.

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다음을 시도하고 넥타이로도 작업했습니다.

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

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Might the UNION operator work for you? Have one SELECT for each section, then UNION them together. Guess it would only work for a fixed number of sections though.

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Q) Finding TOP X records from each group(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

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  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

6 rows selected.

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If you want to produce output grouped by section, displaying only the top n records from each section something like this:

SECTION     SUBSECTION

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

...then the following should work pretty generically with all SQL databases. If you want the top 10, just change the 2 to a 10 toward the end of the query.

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;

To set up:

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';

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While the question was about SQL Server 2005, most people have moved on and if they do find this question, what could be the preferred answer in other situations is one using CROSS APPLY as illustrated in this blog post.

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

This query involves 2 tables. The OP's query only involves 1 table, in case of which a window function based solution might be more efficient.

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You can try this approach. This query returns 10 most populated cities for each country.

   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;

참고URL : https://stackoverflow.com/questions/176964/select-top-10-records-for-each-category