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MySQL获取所有分类的前N条记录_MySQL

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Release: 2016-06-01 13:00:35
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比如有文章表 Article(Id,Category,InsertDate),现在要用SQL找出每种类型中时间最新的前N个数据组成的集合,一段不错的代码,留存备用

SELECT A1.* 
FROM Article AS A1 
   INNER JOIN (SELECT A.Category,A.InsertDate 
         FROM Article AS A 
           LEFT JOIN Article AS B 
            ON A.Category = B.Category 
              AND A.InsertDate <= B.InsertDate 
         GROUP BY A.Category,A.InsertDate 
         HAVING COUNT(B.InsertDate) <= @N 
  ) AS B1 
  ON A1.Category = B1.Category 
    AND A1.InsertDate = B1.InsertDate 
ORDER BY A1.Category,A1.InsertDate DESC
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@N 就是你要取多少条

下面是我用到了一个产品分类表中,superId是大分类,prcid是产品分类。能用SQL完成的功能就要尽量用SQL语句来完成,这既简洁又高效。

SELECT
  A1.*
FROM
  prcKx AS A1
INNER JOIN (
  SELECT
    A.superId,
    A.prcid
  FROM
    prcKx AS A
  LEFT JOIN prcKx AS B ON A.superId = B.superId
  AND A.prcid <= B.prcid
  GROUP BY
    A.superId,
    A.prcid
  HAVING
    COUNT(B.prcid) <= 7
) AS B1 ON A1.superId = B1.superId
AND A1.prcid = B1.prcid
ORDER BY
  superId,
  prcid
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需求是这样的(CSDN上的一个问题):mysql中有个表:article(字段:id,type,date),type有1-10,10种类型。现在要用SQL找出每种类型中时间最新的前N个数据组成的集合。

这个问题应该有很多方法可以实现,下面就来说说在网上看到的一位高手的实现(用一条SQL语句实现的,个人感觉非常好,所以拿来和大家分享):

select a1.* from article a1
inner join
(select a.type,a.date from article a left join article b
on a.type=b.type and a.date<=b.date 
group by a.type,a.date
having count(b.date)<=2
)b1
on a1.type=b1.type and a1.date=b1.date
order by a1.type,a1.date desc
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注:上面sql语句中的2代表的就是前面提到的N。

以上所述就是本文的全部内容了,希望大家能够喜欢。

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