用于查找无效电子邮件地址的Sql脚本

Question

数据导入是从访问数据库完成的,电子邮件地址字段没有验证.有没有人有一个可以返回无效电子邮件地址列表的sql脚本(缺少@等).

谢谢!

Answer 1

SELECT * FROM people WHERE email NOT LIKE '%_@__%.__%'

任何更复杂的东西都可能会返回漏报并且运行得更慢.

在代码中验证电子邮件地址几乎是不可能的.

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Answer 2

这是一个快速简便的解决方案:

CREATE FUNCTION dbo.vaValidEmail(@EMAIL varchar(100))

RETURNS bit as
BEGIN     
  DECLARE @bitRetVal as Bit
  IF (@EMAIL <> '' AND @EMAIL NOT LIKE '_%@__%.__%')
     SET @bitRetVal = 0  -- Invalid
  ELSE 
    SET @bitRetVal = 1   -- Valid
  RETURN @bitRetVal
END 

然后,您可以使用以下函数查找所有行:

SELECT * FROM users WHERE dbo.vaValidEmail(email) = 0

如果您对在数据库中创建函数不满意,可以直接在查询中使用LIKE子句:

SELECT * FROM users WHERE email NOT LIKE '_%@__%.__%'

资源

Answer 3

我发现这个简单的T-SQL查询对于返回有效的电子邮件地址非常有用

SELECT email
FROM People
WHERE email LIKE '%_@__%.__%' 
    AND PATINDEX('%[^a-z,0-9,@,.,_]%', REPLACE(email, '-', 'a')) = 0

PATINDEX位删除所有包含不在允许的az,0-9,'@','.','_'和' - '字符集中的字符的电子邮件地址.

它可以反过来做你想要的这样:

SELECT email
FROM People
WHERE NOT (email LIKE '%_@__%.__%' 
    AND PATINDEX('%[^a-z,0-9,@,.,_]%', REPLACE(email, '-', 'a')) = 0)

Answer 4

select
    email 
from loginuser where
patindex ('%[ &'',":;!+=\/()<>]*%', email) > 0  -- Invalid characters
or patindex ('[@.-_]%', email) > 0   -- Valid but cannot be starting character
or patindex ('%[@.-_]', email) > 0   -- Valid but cannot be ending character
or email not like '%@%.%'   -- Must contain at least one @ and one .
or email like '%..%'        -- Cannot have two periods in a row
or email like '%@%@%'       -- Cannot have two @ anywhere
or email like '%.@%' or email like '%@.%' -- Cant have @ and . next to each other
or email like '%.cm' or email like '%.co' -- Unlikely. Probably typos 
or email like '%.or' or email like '%.ne' -- Missing last letter

这对我有用。必须应用 rtrim 和 ltrim 以避免误报。

来源：http : //sevenwires.blogspot.com/2008/09/sql-how-to-find-invalid-email-in-sql.html

Postgres 版本：

select user_guid, user_guid email_address, creation_date, email_verified, active
from user_data where
length(substring (email_address from '%[ &'',":;!+=\/()<>]%')) > 0  -- Invalid characters
or length(substring (email_address from '[@.-_]%')) > 0   -- Valid but cannot be starting character
or length(substring (email_address from '%[@.-_]')) > 0   -- Valid but cannot be ending character
or email_address not like '%@%.%'   -- Must contain at least one @ and one .
or email_address like '%..%'        -- Cannot have two periods in a row
or email_address like '%@%@%'       -- Cannot have two @ anywhere
or email_address like '%.@%' or email_address like '%@.%' -- Cant have @ and . next to each other
or email_address like '%.cm' or email_address like '%.co' -- Unlikely. Probably typos 
or email_address like '%.or' or email_address like '%.ne' -- Missing last letter
;