Question

有一个scores_score表，其中包含以下列：

id, player_name, value, created_at

我必须获得N（100）个最佳分数，其中：

player_name在结果中必须是唯一的
仅返回给定player_name的最佳分数
结果必须按日期范围过滤

假设我有以下数据：

id      player_name    value          date
1       A               400        2016-09-10
2       B               200        2016-09-12
3       C               400        2016-09-15
4       C               500        2016-09-14
5       B               100        2016-09-20
6       A               6000       2015-01-01
7       B               1200       2016-09-29

想要在2016-09-01和2016-09-20之间获得最佳分数。我应该得到：

id      player_name    value          date
4       C               500        2016-09-14
1       A               400        2016-09-10
2       B               200        2016-09-12

这是我解决它的方法，但是在嵌套的SELECT中存在一个问题，因为它在日期范围内获取整体播放器的最佳分数。

SELECT b.*, a.*
FROM (SELECT player_name, max(value) AS max_value
      FROM scores_score
      GROUP BY player_name
      ORDER BY max(value) DESC) a
INNER JOIN scores_score b ON a.player_name = b.player_name AND a.max_value = b.value
WHERE CAST(b.created_at AS DATE) >= %(date_border)s
ORDER BY b.value DESC
LIMIT 100

Answer 1

distinct on

select *
from (
    select distinct on (player_name) *
    from scores_score
    where date between '2016-09-01' and '2016-09-20'
    order by player_name, value desc
) s
order by value desc
limit 100

Answer 2

这将起作用，并将为您提供预期的输出。使用row_number()窗口函数标记日期之间每个玩家的最高分（rn = 1），然后按value降序排列结果集，最后将输出限制为最高100。

select 
  id, player_name, value, created_at
from (
  select
    id, player_name, value, created_at,
    row_number() over (partition by player_name order by value desc, id) as rn
  from scores_score
  where created_at between '2016-09-01' and '2016-09-20'
  ) ranks
where rn = 1
order by value desc
limit 100

请注意，用于在id函数内进行排序的附加列row_number用于解析关系（即使它在分区内每行只分配一个值），这将涉及具有相同值的两行的同一玩家在给定日期内。这将获得较旧的记录，如果它们与created_at日期不同，您会看到输出的差异： - ）

Answer 3

这个有点麻烦，但应该有用。首先选择日期范围内的玩家和值（a）。然后按球员（b）选择最高分数。然后加入id和日期（c）：

SELECT c.id, c.player_name, c.value, c.date
FROM
scores_score c
INNER JOIN
(SELECT player_name, max(value)
FROM
(SELECT player_name, value
FROM scores_score
WHERE date BETWEEN '2016-09-01' AND '2016-09-20') a
GROUP BY player_name) b
ON c.player_name = b.player_name
AND c.value = b.value
ORDER BY value
LIMIT 100

在这里测试：http://sqlfiddle.com/#!9/10db42/6

在给定日期之间获得最高分数

3 个答案: