主要按照日期获得最新的数据；

今天记录两种方式，并不涉及效率等其他方面问题：

第一种， 利用GROUP BY原理：

select * from (

select * from `authenticationrecord` order by `authenticationtime` desc

) `temp` group by merchantrid order by `authenticationtime` desc

第二种，利用mysql内置函数

select * from `test` where id in(select SUBSTRING_INDEX(group_concat(id order by `date` desc),',',1) from `test` group by category_id ) order by `date` desc

<<<<<<<><>

试验脚本(自己玩)：

create table STAFF(

id int,

sname varchar(50),

dept varchar(50),

salary int,

edlevel int,

hiredate datetime

)

insert into STAFF values('1','张三','开发部','2000','3','2009-10-11');

insert into STAFF values('2','李四','开发部','2500','3','2009-10-01');

insert into STAFF values('3','王五','设计部','2600','5','2010-10-02');

insert into STAFF values('4','王六','设计部','2300','4','2010-10-03');

insert into STAFF values('5','马七','设计部','2100','4','2010-10-06');

insert into STAFF values('6','赵八','销售部','3000','5','2010-10-05');

insert into STAFF values('7','钱九','销售部','3100','7','2010-10-07');

insert into STAFF values('8','孙十','销售部','3500','7','2010-10-06');

select group_concat(id order by hiredate desc) as test from staff

group by dept;

select * from STAFF;

select * from STAFF

group by dept

参考网站：http://blog.csdn.net/agileclipse/article/details/12184635