1. 查找反复的电子邮箱
本人想法（谬误）
SELECT a.Email
FROM Person a,Person b
WHERE 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 Email
FROM Person
GROUP BY Email
HAVING COUNT(Email)>1解二：创立长期表
SELECT Email
FROM(SELECT Email,COUNT(Email) as num
FROM Person
GROUP BY Email
) tmp_table~~~~
WHERE num>1