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。
