Vipin

子查询解决方案 fiddle Link

select * from messages where id in
(select max(id) from messages group by Name)

按连接条件求解 fiddle link

select m1.* from messages m1 
left outer join messages m2 
on ( m1.id<m2.id and m1.name=m2.name )
where m2.id is null

这篇文章的原因是只给小提琴链接。 其他答案中已经提供了相同的sql。