第一种解法: select distinct p1.Email as Email from Person p1, Person p2 where p1.Email=p2.Email and p1.Id>p2.id; 第二种解法: select Email from Person group by Email having count(Email)>1; sql 中有一系列 聚合函数: sum, count, max, avg, 这些函数作用域多条记录上 select sum(populatio…