4

我想选择每个成员的最后一行。

ID     UID      POINT        DATE           TIME

1       1         5       2012-11-29      11:29:03    
2       2        10       2012-11-29      11:38:12    
3       1        10       2012-12-02      05:15:01    
4       3         5       2012-12-02      09:51:34    
5       2         5       2012-12-02      12:14:14    
6       3         5       2012-12-04      12:18:30
7       1         5       2012-12-05      06:00:51

所以我想选择ID，UID和POINT，其中点是每个用户的最高点。结果应该是：

ID     UID      POINT        DATE           TIME

2       2        10       2012-11-29      11:38:12    
3       1        10       2012-12-02      05:15:01      
6       3         5       2012-12-04      12:18:30

我试过这个：

SELECT distinct uid, point, id FROM `test` 
GROUP By uid ORDER BY date DESC, time DESC

和

SELECT id, uid, point FROM `test` 
GROUP BY uid ORDER BY date DESC, time DESC

但我得到了一些错误的结果：

4(3), 2(2), 1(1)
4

8 回答 8

3

尝试：

SELECT id, uid, MAX(point) FROM `test` GROUP BY uid ORDER BY date DESC, time DESC
于 2012-12-20T09:38:53.447 回答
3

此查询将为每个用户选择最高点：

select uid, max(`points`)
from members
group by uid

这将选择用户拥有最大点数的最大 id：

select uid, max(id)
from members
where (uid, `points`) in (select uid, max(`points`)
                          from members
                          group by uid)
group by uid

这是您需要的最终查询：

select members.*
from members
where (uid, id) in (
  select uid, max(id)
  from members
  where (uid, `points`) in (select uid, max(`points`)
                            from members
                            group by uid)
  group by uid)

这表明：

ID  UID  POINT  DATE        TIME
2   2    10     2012-11-29  11:38:12    
3   1    10     2012-12-02  05:15:01      
6   3    5      2012-12-04  12:18:30

这也将给出相同的结果，并且看起来更简单：

SELECT s.*
FROM
  (SELECT members.*
   FROM members
   ORDER BY uid, points desc, id desc) s
GROUP BY uid

我认为它会一直有效，但它没有记录在案！

对最后一个查询的一点解释：MySql 允许您通过查询选择组中的非聚合字段。这里我们分组uid但选择所有列：文档说非聚合列的值将是不确定的（它可以是组内的任何值）但实际上 MySql 只返回第一个遇到的值。而且由于我们在有序子查询上应用了带有非聚合列的 group by，所以第一个遇到的值就是您所需要的。

于 2012-12-20T09:46:50.080 回答
1

试试这个 ：

SELECT id, uid, point FROM `test` 
GROUP BY uid 
ORDER BY point DESC, date DESC, time DESC
于 2012-12-20T09:36:37.467 回答
1

询问：

SQLFIDDLE示例

SELECT t1.*
FROM Table1 t1 
WHERE t1.ID = (SELECT MAX(t3.ID)
               FROM Table1 t3
               WHERE t1.UID=t3.UID
               AND t3.POINT=(SELECT  MAX(t2.POINT)
                               FROM Table1 t2
                               WHERE t2.UID = t3.UID))

结果：

| ID | UID | POINT |                            DATE |     TIME |
-----------------------------------------------------------------
|  2 |   2 |    10 | November, 29 2012 00:00:00+0000 | 11:38:12 |
|  3 |   1 |    10 | December, 02 2012 00:00:00+0000 | 05:15:01 |
|  6 |   3 |     5 | December, 04 2012 00:00:00+0000 | 12:18:30 |
于 2012-12-20T10:04:40.263 回答
0

这是正确的：

SELECT ID, UID, MAX(POINT) FROM `test` GROUP BY UID ORDER BY DATE DESC, TIME DESC
于 2012-12-20T09:46:42.743 回答
0

询问：

SELECT x.*
FROM (SELECT p.*
FROM points p
ORDER BY uid, point desc, id desc) as x
GROUP BY x.uid
;

结果：

ID  UID     POINT   DATE                                TIME
3   1       10      December, 02 2012 00:00:00+0000     January, 01 1970 05:15:01+0000
2   2       10      November, 29 2012 00:00:00+0000     January, 01 1970 11:38:12+0000
6   3       5       December, 04 2012 00:00:00+0000     January, 01 1970 12:18:30+0000
于 2012-12-20T09:47:41.050 回答
0

就这么简单：

SELECT * FROM (SELECT * FROM `points` ORDER BY point DESC) AS `t1` GROUP BY uid;
于 2012-12-20T10:08:11.180 回答
0

SELECT max(point), id, uid FROM test GROUP BY uid ORDER BY point DESC

以这种方式使用，它将解决您的问题。

于 2012-12-20T10:27:04.577 回答