解法(1)：思路为先查询出查询出重复的ID并且取最小值

 select min(Id) Id,Email from Person group by Email

或者

Select min(Id) as Id,distinct Email from Person

然后删除不在ID为此里面的值

delete from Person where Id not in(select Id from ( select min(Id) Id,Email from Person group by Email)t)

解法(2):使用 查询出Email相等ID不是最小的

select B.Id as Id from Person as A,Person as B where A.Id<B.Id and A.Email=B.Email

然后删除查询出来的结果

delete from Person where  Id  in(select Id from(select B.Id as Id from Person as A,Person as B where A.Id<B.Id and A.Email=B.Email) t)