1

我有一个 MySql 表，其中包含如下示例数据：

+---------+---------+--------+---------------------+
|      id | user_id | scores |          created_at |
+---------+---------+--------+---------------------+
|       1 |       1 |     10 | 2012-12-14 02:40:37 |
|       2 |       1 |     20 | 2012-12-14 02:55:54 |
|       3 |       1 |     10 | 2012-12-14 01:17:21 |
|       4 |       2 |     30 | 2012-12-13 01:54:47 |
|       5 |       2 |     55 | 2012-12-15 00:34:39 |
|       6 |       2 |     10 | 2012-12-14 00:20:21 |
+---------+---------+--------+---------------------+

我需要查询它，以便按用户和每小时计算分数。这里假设一小时是created_at跳过的分钟和秒（04:00:00 到 04:59:59 等）。所以是这样的：

+---------+--------+---------------------+
| user_id | scores |          created_at |
+---------+--------+---------------------+
|       1 |     30 | 2012-12-14 02:00:00 |
|       1 |     10 | 2012-12-14 01:00:00 |
|       2 |     30 | 2012-12-13 01:00:00 |
|       2 |     55 | 2012-12-15 00:00:00 |
|       2 |     10 | 2012-12-14 00:00:00 |
+---------+--------+---------------------+

在这个示例数据中，只有第一个用户在一小时内（2012-12-14 02:00:00）玩了不止一次 - 所以他在那一小时内的分数被总结了。

从总分中，我只需要每个用户的最高分（创建排名）。所以最终的预期结果应该是：

+---------+---------------------+---------------------+
| user_id | top_scores_per_hour |                hour |
+---------+---------------------+---------------------+
|       1 |                  30 | 2012-12-14 02:00:00 |
|       2 |                  55 | 2012-12-15 00:00:00 |
+---------+---------------------+---------------------+

我有一个想法如何做这部分......我可以处理数据库之外的休息，但我真的想知道 - 我怎么能用 SQL 查询这个？

4

3 回答 3

1 
SELECT a.user_id,
       a.totalScores top_scores_per_hour,
       a.newTime hour
FROM
    (
      SELECT  user_id, 
              SUM(Scores) totalScores, 
              DATE_Format(created_at, '%Y-%m-%d %H:00:00') newTime
      FROM    TableName 
      GROUP BY user_id, DATE_Format(created_at, '%y-%m-%d %H:00:00')
    ) a INNER JOIN 
    (
      SELECT  user_id, 
              max(DATE_Format(created_at, '%Y-%m-%d %H:00:00')) newTime
      FROM    TableName 
      GROUP BY user_ID
    ) b ON a.user_ID = b.user_ID AND
           a.newTime = b.newTime

输出

╔═════════╦═════════════════════╦═════════════════════╗
║ USER_ID ║ TOP_SCORES_PER_HOUR ║        HOUR         ║
╠═════════╬═════════════════════╬═════════════════════╣
║       1 ║                  30 ║ 2012-12-14 02:00:00 ║
║       2 ║                  55 ║ 2012-12-15 00:00:00 ║
╚═════════╩═════════════════════╩═════════════════════╝
于 2012-12-14T02:38:18.777 回答
1

试试这个：

SELECT user_id, score, createdDate 
FROM (SELECT user_id, SUM(scores) score, DATE_FORMAT(created_at, '%Y-%m-%d %H:00:00') createdDate 
      FROM tablename 
      GROUP BY user_id, createdDate
      ORDER BY user_id, score DESC) AS A 
GROUP BY user_id
于 2012-12-14T09:34:28.060 回答
0

怎么样

Select user_id, Max(totalHouirlyScore) TopHourlyScore
From (Select user_id, Sum(scores) totalHouirlyScore
      From tablename
      Group By user_id, 
            DateFormat(created_at, '%y%m%d%H')) Z
于 2012-12-14T02:25:39.237 回答