Question

表架构设置如下：

userID      Points       timestamp

1           40
3           20
1           10
4           15
3           5

需要能够提供显示以下内容的报告：

Total Points Allocated for the Day (0 if none allocated), (nice to have) To what userID's the points were allocated to for that day

我尝试了以下内容，因为您可以看到不正确

SELECT uid, DATE(time_stamp) AS date, SUM(points) AS total_points
FROM table
GROUP BY date

Answer 1

假设您有值作为时间戳，我将使用列名userid，points和time_stamp以及用户点的表名：

SELECT userID,
       sum(points),
       date(timestamp) as date
FROM   userpoints
GROUP BY userID, date(timestamp)

所以：

userID      Points       timestamp 

1           40           18-8-2010 12:00:00.000
3           20           18-8-2010 12:00:00.000
1           10           18-8-2010 12:00:00.000
4           15           18-8-2010 12:00:00.000
3           5            18-8-2010 12:00:00.000

会导致：

userid      points       date
1           50           18-8-2010
3           25           18-8-2010
4           15           18-8-2010

更新：针对UNIX问题进行了优化

SELECT userID,
       sum(points),
       from_unixtime(timestamp, 'DD-MM-YYYY') as date
FROM   userpoints
GROUP BY userID, date(timestamp)

Answer 2

MySQL没有递归功能，所以你可以使用NUMBERS表技巧来获取一个天数列表，以便在看到零关系归属的日子之前联系起来。

创建一个只保存递增数字的表 - 使用auto_increment很容易做到：

DROP TABLE IF EXISTS `example`.`numbers`;
CREATE TABLE  `example`.`numbers` (
  `id` int(10) unsigned NOT NULL auto_increment,
   PRIMARY KEY  (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

使用以下方法填充表格：
```
INSERT INTO NUMBERS
  (id)
VALUES
  (NULL)
```
...根据需要提供尽可能多的价值。

使用DATE_ADD构建一个时间列表，根据NUMBERS.id值增加月份：

SELECT x.*
  FROM (SELECT DATE_ADD('2010-01-01', INTERVAL n.id - 1 DAY)
          FROM numbers n) x

根据日期时间部分LEFT JOIN到您的数据表：

   SELECT up.userid,
          COALESCE(SUM(up.points), 0) AS points
          DATE(x.dt) AS date
     FROM (SELECT DATE_ADD('2010-01-01', INTERVAL n.id - 1 MONTH) AS dt
             FROM numbers n) x
LEFT JOIN USERPOINTS up ON DATE(FROM_UNIXTIME(up.timestamp)) = DATE(x.dt)
 GROUP BY up.userid, DATE(x.dt)

Answer 3

SELECT DATE(time_stamp) as date1, userID, SUM(Points) as total_points
FROM table1
GROUP BY date1, userID
WITH ROLLUP

具有null userID而非null date1的记录显示在该日期分配了多少个点，具有null userID和null date1的记录 - 分配的总分数。

需要有关MYSQL查询分组的帮助

3 个答案: