解法(1)：思路為先查詢出查詢出重復的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

然后刪除不在ID為此里面的值

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 查詢出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

然后刪除查詢出來的結果

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)

總結

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