获取每个客户的前12条记录 [英] fetching top 12 records of each client

问题描述

我在表中具有每个客户端的性能数据

有字段

clientid(varchar(20))，date(datetime)，性能(数字)

现在我想要的是，根据他们的表现来获取每位客户的前12条12条记录.

请帮忙..

在此先感谢

I am having performance data of each client in a table

having fields

clientid(varchar(20)),date(datetime), performance (number)

now what I want is, to fetch top 12 twelve record of each client on the basis of their performance.

please help..

thanks in advance

推荐答案

尝试一下:

Try this:

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(*) <= 12;

经过测试的查询

Tested query

SELECT DISTINCT TOP 12(Clientid), MAX(Performance) AS MAX_PerformanceOfClient FROM [TableName]
GROUP BY Clientid
ORDER BY MAX(Performance) DESC

尝试以下语句

Try this statement

select distinct top 12 (clientid), performance from tableName
Order By performance desc

问候，
T

Regards,
T

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