找到每个日期完成的最大值[关闭]

Question

请原谅标题，但我完全不知道如何描述这个。

我有一张这样的桌子：

trackID playerID score   date
1        2       4510    1494075555
1        2       4507    1494076300
1        2       4513    1494076561
2        3       39455   1494083772
3        3       5665    1494089018
2        2       38444   1494074519
4        3       34443   1494089138
5        3       56443   1494260918

我只想要每首曲目的第一次完成，一个曲目可以有多个玩家在同一日期完成。 我想在一天内完成最多的第一次完成。

在上表中，玩家 3 在 2017 年 5 月 6 日首次完成了 3 张地图。 在 8.5.2017 上，他只完成了一首曲目。我只想包括他完成最多曲目的日期。

我想要的结果：

playerID trackID count(trackID)  date
3        2,3,4       3           6.5.2017
2        1,2         2           6.5.2017

Answer 1

这很丑陋，但我们能做什么。 增加样本量以测试有些复杂的算法并显示限制

所以为了更好地验证我的结果。

桌子

CREATE TABLE table1
    (`trackID` int, `playerID` int, `score` int, `date` int)
;

INSERT INTO table1
    (`trackID`, `playerID`, `score`, `date`)
VALUES
    (1, 2, 4510, 1494075555),
    (1, 2, 4507, 1494076300),
    (1, 2, 4513, 1494076561),
    (2, 3, 39455, 1494083772),
    (3, 3, 5665, 1494089018),
    (2, 2, 38444, 1494074519),
    (4, 3, 34443, 1494089138),
    (5, 3, 56443, 1494260918),
    (6, 3, 56443, 1494260928),
    (7, 3, 56443, 1494260938),
    (1, 4, 4510,1494324530),
    (2, 4, 4510,1494324530),
    (3, 4, 4510,1494324530),
    (4, 4, 4510,1494324530),
    (5, 4, 4510,1494324530),
    (6, 4, 4510,1494324530)
;

有了这个选择语句

SELECT
  a4.trackID, a4.playerID, 
  MAX(
  a4.countgroup
  ) countgroup,
  MAX(
  CONCAT(day_a,".",month_a,".",year_a)
  )  `date`
FROM
(
   SELECT 
     Group_COnCAT(DISTINCT trackID) trackID,
     playerID
     , Count(*) countgroup 
     ,YEAR(FROM_UNIXTIME(`date`)) year_a
     ,MONTH(FROM_UNIXTIME(`date`))  month_a
     ,DAY(FROM_UNIXTIME(`date`))  day_a
   FROM
    (Select 
       DISTINCT trackID, playerID,MAX(`date`) `date`
     FROM table1
    GROUP by  playerID,trackID
    ) c1
  GROUP BY playerID,YEAR(FROM_UNIXTIME(`date`)) 
         ,MONTH(FROM_UNIXTIME(`date`)) ,DAY(FROM_UNIXTIME(`date`))
 ) a4
inner join
( SELECT
   a2.playerID, a2.countgroup,a2.`date`
 FROM
 (
   SELECT playerID,  Count(*) countgroup, MAX(`date`) `date`
   FROM
    (
     Select 
       DISTINCT trackID, playerID,MAX(`date`) `date`
     FROM table1
     GROUP by  playerID,trackID
    ) c1
    GROUP BY playerID,YEAR(FROM_UNIXTIME(`date`)) 
         ,MONTH(FROM_UNIXTIME(`date`)) ,DAY(FROM_UNIXTIME(`date`))
  ) a2 
   INNER JOIN
  (
  SELECT playerID,max(countgroup) maxcountgroup, MAX( `date`)  `date`
   FROM
   (
     SELECT playerID,  Count(*) countgroup, MAX( `date`)  `date`
      FROM
      (
        Select 
          DISTINCT trackID, playerID,MAX(`date`) `date`
        FROM table1
        GROUP by  playerID,trackID
      ) c1
      GROUP BY playerID,YEAR(FROM_UNIXTIME(`date`)) 
         ,MONTH(FROM_UNIXTIME(`date`)) ,DAY(FROM_UNIXTIME(`date`))
    ) g1  
     GROUP BY playerID,YEAR(FROM_UNIXTIME(`date`)) 
         ,MONTH(FROM_UNIXTIME(`date`)) ,DAY(FROM_UNIXTIME(`date`))
  ) a3
  ON a2.playerID = a3.playerID AND a3.maxcountgroup = a2.countgroup
    and YEAR(FROM_UNIXTIME(a2.`date`))  = YEAR(FROM_UNIXTIME(a3.`date`)) 
     #and MONTH(FROM_UNIXTIME(a2.`date`)) = MONTH(FROM_UNIXTIME(a3.`date`))
     #and DAY(FROM_UNIXTIME(a2.`date`)) = DAY(FROM_UNIXTIME(a3.`date`))
) b4
ON a4.playerID = b4.playerID AND a4.countgroup = b4.countgroup
GROUP BY playerID,trackID;

你会得到以下结果

trackID         playerID    countgroup  date
1,2             2           2           6.5.2017
2,3,4           3           3           6.5.2017
5,6,7           3           3           8.5.2017
1,2,3,4,5,6     4           6           9.5.2017

这很复杂： 所以在表 a4 中我得到了所需的数据，但是每个玩家都有多天，我必须选择赢得最多曲目的日子。

进入表 b4，它为每个玩家选择到达的日期和最多的曲目。

您在示例中看到的问题是，当有两天或更多天征服了相同数量的轨道时。

这是 dbfiddle 示例，以便您对其进行测试 https://dbfiddle.uk/?rdbms=mysql_5.7&fiddle=88bb4f9c208ccf6a9865b79b1681faa5

edit2：必须在 SEELCT 和 b4 的 on 子句中添加日期列