Question

我有两个表players和games，创建如下：

CREATE TABLE IF NOT EXISTS `players` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(255) NOT NULL,
  `created_at` datetime NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;

CREATE TABLE IF NOT EXISTS `games` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `player` int(11) NOT NULL,
  `played_at` datetime NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;

我希望每天提取3个值：

当天创建的玩家数量
当天玩的球员数量
当天第一次参加比赛的球员数量

因此，假设玩家表格如下所示：

+----+--------+---------------------+
| id | name   | created_at          |
+----+--------+---------------------+
|  1 | Alan   | 2016-02-01 00:00:00 |
|  2 | Benny  | 2016-02-01 06:00:00 |
|  3 | Calvin | 2016-02-02 00:00:00 |
|  4 | Dan    | 2016-02-03 00:00:00 |
+----+--------+---------------------+

游戏表如下：

+----+--------+---------------------+
| id | player | played_at           |
+----+--------+---------------------+
|  1 |      1 | 2016-02-01 01:00:00 |
|  2 |      3 | 2016-02-02 02:00:00 |
|  3 |      2 | 2016-02-03 14:00:00 |
|  4 |      3 | 2016-02-03 17:00:00 |
|  5 |      3 | 2016-02-03 18:00:00 |
+----+--------+---------------------+

然后查询应返回类似

+------------+-----+--------+-------+
| day        | new | played | first |
+------------+-----+--------+-------+
| 2016-02-01 | 2   | 1      | 1     |
| 2016-02-02 | 1   | 1      | 1     |
| 2016-02-03 | 1   | 2      | 1     |
+------------+-----+--------+-------+

我有一个解决方案1（ new ）：

SELECT Date(created_at) AS day,
       Count(*)         AS new
FROM   players
GROUP  BY day;

这很容易。我想我也有2个解决方案（播放），感谢MySQL COUNT DISTINCT：

select Date(played_at) AS day,
       Count(Distinct player) AS played
FROM   games
GROUP  BY day;

但我不知道如何获得3（ first ）所需的结果。我也不知道如何将所有内容放在一个查询中，以节省执行时间（games表可能包含数百万条记录）。

如果您需要，这是一个插入示例数据的查询：

INSERT INTO `players` (`id`, `name`, `created_at`) VALUES
(1, 'Alan', '2016-02-01 00:00:00'),
(2, 'Benny', '2016-02-01 06:00:00'),
(3, 'Calvin', '2016-02-02 00:00:00'),
(4, 'Dan', '2016-02-03 00:00:00');

INSERT INTO `games` (`id`, `player`, `played_at`) VALUES
(1, 1, '2016-02-01 01:00:00'),
(2, 3, '2016-02-02 02:00:00'),
(3, 2, '2016-02-03 14:00:00'),
(4, 3, '2016-02-03 17:00:00'),
(5, 3, '2016-02-03 18:00:00');

Answer 1

一个版本是将所有相关数据导入联合并从那里进行分析;

SELECT SUM(type='P') new, 
       COUNT(DISTINCT CASE WHEN type='G' THEN pid END) played, 
       SUM(type='F') first 
FROM (
  SELECT id pid, DATE(created_at) date, 'P' type FROM players 
  UNION ALL 
  SELECT player, DATE(played_at) date,  'G' FROM games 
  UNION ALL 
  SELECT player, MIN(DATE(played_at)),  'F' FROM games GROUP BY player
) z 
GROUP BY date;

在工会中;

类型为P的记录是玩家创建统计信息类型为G的记录是与玩家相关的游戏统计数据类型为F的记录是玩家第一次玩游戏时的统计数据。

Answer 2

您可以根据min（playing_at）计算临时表的结果，并通过

进行过滤

select count(player) from 
   (  select player, min(played_at)  
      from games 
      group by player 
      having min(played_at) = YOUR_GIVEN_DATE ) as t;

Answer 3

此查询将为您提供结果：

select day,( select count(distinct(id)) from players where Date(created_at) = temp.day ) as no_created_at ,
( select count(distinct(player)) from games where Date(played_at) = temp.day) as no_played_at,
( select count(distinct(player)) from games  where Date(played_at) = 
(select min(Date(played_at)) from games internal_games 
where internal_games.player =games.player and Date(games.played_at) = temp.day )) as no_first_played_at
 from (
SELECT Date(created_at) AS day     
FROM   players
GROUP  BY day 
union 
select Date(played_at) AS day
FROM   games
GROUP  BY day) temp

和输出：

Answer 4

这是一个带有一堆子查询的解决方案，它解释了玩家可能在没有游戏的情况下创建的可能性，反之亦然：

select
    all_dates.date as day,
    ifnull(new.num, 0) as new,
    ifnull(players.num, 0) as players,
    ifnull(first.num, 0) as first
from (
    select date(created_at) as date from players
    union
    select date(played_at) from games
) as all_dates
left join (
    select date(created_at) as created_at_date, count(*) as num
    from players
    group by created_at_date
) as new on all_dates.date = new.created_at_date
left join (
    select date(played_at) as played_at_date, count(distinct player) as num
    from games
    group by played_at_date
) as players on all_dates.date = players.played_at_date
left join (
    select min_date, count(*) num
    from (
        select player, date(min(played_at)) as min_date
        from games
        group by player
    ) as players_first
    group by min_date
) as first on all_dates.date = first.min_date
order by day;

计算具有在给定日期首先出现的相关记录的记录

4 个答案: