【问题标题】:How to get maximum row of each group?如何获得每组的最大行数?
【发布时间】:2013-01-25 00:30:05
【问题描述】:

我有一个数据库,人们可以在其中为不同地方的人提交选票。在给定的时间,我想找出每个地方谁的票数最多。 (一个人可以在两个不同的地方投票)

这是我目前的 SQL:

SELECT placeId, userVotedId, cnt 
FROM 
    (SELECT uvo.userVotedId, p.placeId, count(*) as cnt 
     FROM users as u, users_votes as uvo, places as p 
     WHERE u.userId = uvo.userVotedId 
       AND p.placeId = uvo.placeId 
     GROUP BY userVotedId, placeId) 
AS RESULT

这给了我这个结果:

现在,这些是我真正想要的行:

我的查询中缺少什么,所以我可以得到这个?

  • 我希望每个位置有一个结果。所以我应该只看到不同的 placeId,其中 userVotedId 获得了最多的选票。

  • 如果出现平局,则随机获胜!

【问题讨论】:

  • 你得到的和你想要的是一样的。所以,您的查询有效!
  • @cha 我也这么认为,但关键是要消除双 placeid = 51。可能有更容易掌握的方法来指出这一点,而不是四个凹凸不平的红色框;-)跨度>
  • 如果您尝试在根据结果的 placeID 进行分组后基于 cnt 获取第一行,这比最初出现的问题更难。在 SQL Server 中,我认为您可以使用 APPLY 完成此操作,但在 MySQL 中不支持 (?)。
  • 你测试过我的答案吗——这是小提琴(sqlfiddle.com/#!2/23bd7/6)?也许不是最优雅的,但它可以工作(如果 MySQL 支持 CTE 会更好)。祝你好运!
  • 完成 -- 谢谢,很高兴我能提供帮助。

标签: mysql sql greatest-n-per-group


【解决方案1】:

看来你还需要一个聚合体。在您的 cnt 值和 GROUP BY placeId, userVotedId 上使用 MAX() 聚合:

SELECT placeId, userVotedId, max(cnt)
FROM 
(
  SELECT uvo.userVotedId, p.placeId, count(*) as cnt 
  FROM users as u
  INNER JOIN users_votes as uvo
    ON u.userId = uvo.userVotedId 
  INNER JOIN places as p 
    ON p.placeId = uvo.placeId 
  GROUP BY userVotedId, placeId
) AS RESULT
GROUP BY placeId, userVotedId

注意:我将您的查询更改为使用 JOIN 语法,而不是表之间的逗号。

编辑,根据您的评论,以下应该有效:

select total.uservotedid,
  total.placeid,
  total.cnt
from
(
  SELECT uvo.userVotedId, p.placeId, count(*) as cnt 
  FROM users as u
  INNER JOIN users_votes as uvo
    ON u.userId = uvo.userVotedId 
  INNER JOIN places as p 
    ON p.placeId = uvo.placeId 
  GROUP BY userVotedId, placeId
) total
inner join
(
  select max(cnt) Mx, placeid
  from
  (
    SELECT uvo.userVotedId, p.placeId, count(*) as cnt 
    FROM users as u
    INNER JOIN users_votes as uvo
      ON u.userId = uvo.userVotedId 
    INNER JOIN places as p 
      ON p.placeId = uvo.placeId 
    GROUP BY userVotedId, placeId
  ) mx
  group by placeid
) src
  on total.placeid = src.placeid
  and total.cnt = src.mx

SQL Fiddle with Demo

结果是:

| USERVOTEDID | PLACEID | CNT |
-------------------------------
|          65 |      11 |   1 |
|          67 |      13 |   1 |
|          67 |      25 |   1 |
|          67 |      51 |   2 |

编辑#2,如果你想在平局时返回一个随机数,那么你可以使用用户变量:

select uservotedid,
  placeid, 
  cnt
from
(
  select total.uservotedid,
    total.placeid,
    total.cnt,
    @rownum := case when @prev = total.placeid then @rownum+1 else 1 end rownum,
    @prev := total.placeid pplaceid
  from
  (
    SELECT uvo.userVotedId, p.placeId, count(*) as cnt 
    FROM users as u
    INNER JOIN users_votes as uvo
      ON u.userId = uvo.userVotedId 
    INNER JOIN places as p 
      ON p.placeId = uvo.placeId 
    GROUP BY userVotedId, placeId
  ) total
  inner join
  (
    select max(cnt) Mx, placeid
    from
    (
      SELECT uvo.userVotedId, p.placeId, count(*) as cnt 
      FROM users as u
      INNER JOIN users_votes as uvo
        ON u.userId = uvo.userVotedId 
      INNER JOIN places as p 
        ON p.placeId = uvo.placeId 
      GROUP BY userVotedId, placeId
    ) mx
    group by placeid
  ) src
    on total.placeid = src.placeid
    and total.cnt = src.mx
  order by total.placeid, total.uservotedid
) src
where rownum = 1
order by placeid, uservotedid

SQL Fiddle with Demo

【讨论】:

  • 您只想要按位置的最大值吗?还是您也想要用户 ID?
  • 我希望每个位置有一个结果。所以我应该只看到不同的 placeId,其中 userVotedId 获得了最多的选票。
  • @bluefeet -- 和我的回答一模一样 :-)
  • @sgeddes 我没有注意我正在研究 SQL 小提琴答案的其他答案。
  • 我以为你有,但后来我添加了一个投票,但它没有给我正确的结果:sqlfiddle.com/#!2/fd0b3/1
【解决方案2】:
SELECT placeId, userVotedId, MAX(cnt)
    FROM (SELECT uvo.userVotedId, p.placeId, count(*) AS cnt 
          FROM users as u, users_votes as uvo, places as p 
          WHERE u.userId = uvo.userVotedId AND p.placeId = uvo.placeId 
          GROUP BY userVotedId, placeId) AS RESULT
    GROUP BY  placeId

【讨论】:

  • 不,我之前尝试过(最后按 placeId 分组),但问题是它没有给出正确的答案。看结果:placeId=51的userVotedId应该是67,其实是63。
【解决方案3】:

为简单起见,我将您的查询称为测试:

SELECT * 
FROM Test T JOIN (
SELECT t.placeId, Max(t.cnt) maxcnt
FROM Test t
GROUP BY t.placeId) T2 ON T.placeId = T2.placeId and T.cnt = T2.maxcnt

这里是Fiddle

顺便说一句——测试=:

SELECT uvo.userVotedId, p.placeId, count(*) as cnt 
  FROM users as u
  INNER JOIN users_votes as uvo
    ON u.userId = uvo.userVotedId 
  INNER JOIN places as p 
    ON p.placeId = uvo.placeId 
  GROUP BY userVotedId, placeId

祝你好运。

--EDIT -- 按照要求,这是最终代码:

SELECT * 
FROM (SELECT uvo.userVotedId, p.placeId, count(*) AS cnt 
          FROM users as u, users_votes as uvo, places as p 
          WHERE u.userId = uvo.userVotedId AND p.placeId = uvo.placeId 
          GROUP BY userVotedId, placeId) T JOIN (
SELECT t.placeId, Max(t.cnt) maxcnt
FROM (SELECT uvo.userVotedId, p.placeId, count(*) AS cnt 
          FROM users as u, users_votes as uvo, places as p 
          WHERE u.userId = uvo.userVotedId AND p.placeId = uvo.placeId 
          GROUP BY userVotedId, placeId) t
GROUP BY t.placeId) T2 ON T.placeId = T2.placeId and T.cnt = T2.maxcnt

【讨论】:

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