Select top 10 records for each category

user_14752 0
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I want to return top 10 records from each section in one query. Can anyone help with how to do it? Section is one of the columns in the table.

Database is SQL Server 2005. I want to return the top 10 by date entered. Sections are business, local, and feature. For one particular date I want only the top (10) business rows (most recent entry), the top (10) local rows, and the top (10) features.

greatest-n-per-group sql-server-2005 sql-server sql
14 Answers

If you are using SQL 2005 you can do something like this...

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

If your RankCriteria has ties then you may return more than 10 rows and Matt's solution may be better for you.

user_68190

This works on SQL Server 2005 (edited to reflect your clarification):

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
Matt_Hamilton_6150

I do it this way:

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;

update: This example of GROUP BY works in MySQL and SQLite only, because those databases are more permissive than standard SQL regarding GROUP BY. Most SQL implementations require that all columns in the select-list that aren't part of an aggregate expression are also in the GROUP BY.

Bill_Karwin_208600

If you know what the sections are, you can do:

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
user_3690

I know this thread is a little bit old but I've just bumped into a similar problem (select the newest article from each category) and this is the solution I came up with :

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

This is very similar to Darrel's solution but overcomes the RANK problem that might return more rows than intended.

Diadistis_474010

Tried the following and it worked with ties too.

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
user_34378850

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.

sblundy_48930

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.

user_5217990
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