1. 查找反复的电子邮箱
本人想法(谬误)
SELECT a.EmailFROM Person a,Person bWHERE a.Email = b.Email
后果:表内全副呈现
解决思路:计算每封邮件的存在次数
SELECT EmailFROM Person GROUP BY EmailHAVING COUNT(Email)>1
解二:创立长期表
SELECT EmailFROM( SELECT Email,COUNT(Email) as num FROM Person GROUP BY Email) tmp_table~~~~WHERE num>1