带时间间隔的Sql查询

我想在2014-07-01和2014-08-01之间获得最好的3天用户

有人能帮我吗？我已经被困在这里三天了

在实际分数表中，条目为10:00至22:00，每小时1个条目

每天总共有12个条目，每个玩家有时可能少于1或2个条目

这是我试图获得的输出：

ID | User_ID | Username | Sum(Score) | Date
--------------------------------------------------
 1  |    1   |   Xxx    |     52     | 2014-07-01
 2  |    1   |   Xxx    |     143    | 2014-07-02
 3  |    2   |   Yyy    |     63     | 2014-07-01
...

评分表：

用户表：

---

我认为您需要首先按日期进行聚合，然后使用行数选择前三个。要进行聚合，请执行以下操作：

select s.user_id, sum(s.datetime, 'day') as theday, sum(score) as score,
       row_number() over (partition by s.user_id order by sum(score) desc) as seqnum
from scores s
group by s.user_id;

要获取其余信息，请将其用作子查询或CTE：

select u.*, s.score
from (select s.user_id, sum(s.datetime, 'day') as theday, sum(s.score) as score,
             row_number() over (partition by s.user_id order by sum(s.score) desc) as seqnum
      from scores s
      group by s.user_id
     ) s join
     users u
     on s.user_id = u.users_id
where seqnum <= 3
order by u.user_id, s.score desc;

---

我想你希望第三排的结果是63。你说得对，我把它改成了63。

select s.user_id, sum(s.datetime, 'day') as theday, sum(score) as score,
       row_number() over (partition by s.user_id order by sum(score) desc) as seqnum
from scores s
group by s.user_id;

select u.*, s.score
from (select s.user_id, sum(s.datetime, 'day') as theday, sum(s.score) as score,
             row_number() over (partition by s.user_id order by sum(s.score) desc) as seqnum
      from scores s
      group by s.user_id
     ) s join
     users u
     on s.user_id = u.users_id
where seqnum <= 3
order by u.user_id, s.score desc;

SELECT      'group has no id'   as ID,
            u.ID                as User_ID,
            u.Username,
            sum(s.Score)        "Sum(Score)",
            s.Datetime::date    as Date
FROM        User u,
            Score s
WHERE       u.id = s.User_ID
AND         s.Datetime BETWEEN '2014-07-01' AND '2014-08-01 23:59:59'
GROUP BY    u.ID, u.Username, s.Datetime::date
ORDER BY    sum(s.Score) DESC
LIMIT 3;