关于group by语句mysql
我遇到这样的一种需求：
A表字段：product_id(产品id)，value（产品价格），datetime（时间）
每一种产品每天的价格都是会变化滴，我这边的需求就是取出前十条产品价格，要求时间是最新的十条，产品是不同的十种。
这可难倒我了，使用group by product_id吧，没有办法获得时间，按时间排序吧，无法得到不同的十种产品
求一条SQL语句，实现需求

[解决办法]
按时间牌序然后distinct步行么

[解决办法]
select *
from A a
where not exists (select 1 from A b where a.product_id=B.product_id and A.datetime<B.datetime)
order by A.datetime desc
limit 10;
[解决办法]

SQL code

mysql> create table product_value(    ->  `id` int not null auto_increment,    ->  `product_id` int not null,    ->  `value` decimal(11,2),    ->  `datetime` datetime not null,    ->  primary key(`id`)    -> )engine=innodb default charset=utf8;Query OK, 0 rows affected (0.08 sec)mysql> insert into product_value (product_id,value,datetime)    -> values    -> (1,11.2,'2012-05-25 11:00:05'),    -> (2,21.2,'2012-05-25 11:02:35'),    -> (3,10.3,'2012-05-25 11:15:38'),    -> (1,11.5,'2012-05-25 11:12:40'),    -> (2,19.8,'2012-05-25 12:32:00'),    -> (3,8.15,'2012-05-25 15:12:40'),    -> (4,68.15,'2012-05-25 16:12:40'),    -> (1,15,'2012-05-25 19:12:40'),    -> (3,8.16,'2012-05-25 20:22:42');Query OK, 9 rows affected (0.03 sec)Records: 9  Duplicates: 0  Warnings: 0mysql> select * from product_value;+----+------------+-------+---------------------+| id | product_id | value | datetime            |+----+------------+-------+---------------------+|  1 |          1 | 11.20 | 2012-05-25 11:00:05 ||  2 |          2 | 21.20 | 2012-05-25 11:02:35 ||  3 |          3 | 10.30 | 2012-05-25 11:15:38 ||  4 |          1 | 11.50 | 2012-05-25 11:12:40 ||  5 |          2 | 19.80 | 2012-05-25 12:32:00 ||  6 |          3 |  8.15 | 2012-05-25 15:12:40 ||  7 |          4 | 68.15 | 2012-05-25 16:12:40 ||  8 |          1 | 15.00 | 2012-05-25 19:12:40 ||  9 |          3 |  8.16 | 2012-05-25 20:22:42 |+----+------------+-------+---------------------+9 rows in set (0.00 sec)mysql> select * from (select product_id,max(datetime) as datetime from product_value group by product_id desc limit 10) t order by datetime desc;+------------+---------------------+| product_id | datetime            |+------------+---------------------+|          3 | 2012-05-25 20:22:42 ||          1 | 2012-05-25 19:12:40 ||          4 | 2012-05-25 16:12:40 ||          2 | 2012-05-25 12:32:00 |+------------+---------------------+4 rows in set (0.00 sec)mysql>
[解决办法]
晕，忘记价格了
[解决办法]
SQL code
select * from product_value where id in(select max(id) as id from product_value group by product_id) order by datetime desc limit 10;+----+------------+-------+---------------------+| id | product_id | value | datetime            |+----+------------+-------+---------------------+|  9 |          3 |  8.16 | 2012-05-25 20:22:42 ||  8 |          1 | 15.00 | 2012-05-25 19:12:40 ||  7 |          4 | 68.15 | 2012-05-25 16:12:40 ||  5 |          2 | 19.80 | 2012-05-25 12:32:00 |+----+------------+-------+---------------------+
[解决办法]
SQL code
select *from A twhere not exists (select 1 from A where product_id =t.product_id and  datetime>t.datetime)order by datetime desc limit 10
[解决办法]
参考下贴中的多种方法

http://blog.csdn.net/acmain_chm/article/details/4126306
[征集]分组取最大N条记录方法征集，及散分.... 
 
[解决办法]
select * from A a
where not exists(select 1 from A where product_id=a.product_id and datetime>a.datetime)
order by datatime desc
limit 10 ;