选择最后一行

Question

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 

我试着用这样的：

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) 

Answer 1

这个查询为每个用户选择的最高点：

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

，这将选择用户拥有最大分数的最大ID：

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

，这是你所需要的最终查询：

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) 

显示：

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 

这也会产生相同的结果，看上去简单：

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

我认为它会一直工作，但它'没有记录！

对最后一个查询的一点解释：MySql允许您通过查询来选择组中的非聚集字段。在这里，我们按uid进行分组，但选择所有列：文档说明非聚集列的值将不确定（它可以是组内任意值），但实际上MySql只是返回遇到的第一个值。而且，由于我们在有序子查询中应用了一个非聚集列的组，所以遇到的第一个值就是您需要的值。

Answer 2

试试这个：

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

Answer 3

尝试：

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

Answer 4

这是正确的：

SELECT ID, UID, MAX(POINT) FROM `test` GROUP BY UID ORDER BY DATE DESC, TIME DESC 

Answer 5

• SQLFIDDLE Reference：从第一个答案更新以匹配预期结果。

查询：

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

Reusults：

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 

Answer 6

查询：

SQLFIDDLEEXample

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 | 

Answer 7

就算这样简单：

SELECT * FROM (SELECT * FROM `points` ORDER BY point DESC) AS `t1` GROUP BY uid; 

Answer 8

SELECT MAX（点），ID，UID来自测试 GROUP BY UID ORDER BY点DESC

使用这种方式，它将解决你的问题。