每组获得N个MYSQL [英] Get N per group MYSQL

问题描述

这不是一个相同的问题:

IT IS NOT THE SAME QUESTION THAN : Using LIMIT within GROUP BY to get N results per group?

但是我承认这是相似的.

but i admit it's similar.

我需要选择每个人的前2行. 这些行按收到的年份排序

I need to select the first 2 rows per person. the rows are ordered by Year recieved

问题:同一个月可能输入了2个数据(日期输入为YYYY-MM)

Problem : there is a possibility than 2 data were entered the same month (Date is entered YYYY-MM)

我所伴随的查询(在提到的问题之后)被卡在了BIG循环中.

The query I came with (following the refered question) is stuck in an BIG loop.

SELECT *
FROM `table_data` as b
WHERE (
    SELECT count(*) FROM `table_data` as a
        WHERE a.Nom = b.Nom and a.year < b.year
    ) <= 2;

样本数据:

  A  |   year   |  Nom
---------------------
  b  |  2011-01 | Tim
---------------------
  d  |  2011-01 | Tim
---------------------
  s  |  2011-01 | Tim
---------------------
  a  |  2011-03 | Luc
---------------------
  g  |  2011-01 | Luc
---------------------
  s  |  2011-01 | Luc

应该导出:

  A  |   year   |  Nom
---------------------
  b  |  2011-01 | Tim
---------------------
  d  |  2011-01 | Tim
---------------------
  a  |  2011-03 | Luc
---------------------
  g  |  2011-01 | Luc

推荐答案

(
 -- First get a set of results as if you only wanted the latest entry for each
 -- name - a simple GROUP BY from a derived table with an ORDER BY
  SELECT *
  FROM (
    SELECT *
    FROM `table_data`
    ORDER BY `year` DESC
  ) `a`
  GROUP BY `Nom`
)
UNION
(
 -- Next union it with the set of result you get if you apply the same criteria
 -- and additionally specify that you do not want any of the rows found by the
 -- first operation
  SELECT *
  FROM (
    SELECT *
    FROM `table_data`
    WHERE `id` NOT IN (
      SELECT `id`
      FROM (
        SELECT *
        FROM `table_data`
        ORDER BY `year` DESC
      ) `a`
      GROUP BY `Nom`
    )
    ORDER BY `year` DESC
  ) `b`
  GROUP BY `Nom`
)
 -- Optionally apply ordering to the final results
ORDER BY `Nom` DESC, `year` DESC

我确信这样做的方法更短，但是现在我无法为自己的生活弄清楚它是什么.尽管可以正常工作-假设您有一个主键(应该有)，并且它称为id.

I feel sure there is a shorter way of doing it but right now I can't for the life of me work out what it is. That does work, though - assuming you have a primary key (which you should) and that it is called id.

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