SQL脚本查找无效的电子邮件地址

69

已从访问数据库完成数据导入，并且电子邮件地址字段上没有验证。是否有人拥有可以返回无效电子邮件地址列表（缺少@等）的sql脚本。

sql  validation  email 
— 坎波
source

Answers:

171 
SELECT * FROM people WHERE email NOT LIKE '%_@__%.__%'

任何更复杂的操作都可能返回假阴性并运行得更慢。

用代码验证电子邮件地址实际上是不可能的。

编辑：相关问题

— 托玛拉克
source
5
我已经使用过这个了，它多年来没有让我失望。我认为自己在正则表达式方面非常擅长，但我想一个赛昂人
— 乍得·格兰特2009年
1
已经太复杂和错误了。foo @ bar是一个合法的电子邮件地址（假设存在“ .bar” TLD，并且具有地址或MX记录）。
— bortzmeyer
3
甚至称其为“不太可能”已经很英国了。该表达式不适用于验证电子邮件地址或检查所有极端情况。这是一项基本的健全性检查，涵盖所有情况的99.9％，而不会产生假阴性，我没有另外指出。
— Tomalak
3
注释“太复杂和错误”后跟“太简单”可以很好地总结所有电子邮件验证。这是一个很棒的健全性检查表达，在许多情况下特别有用。
— toxaq 2014年
4
q.com是在美国受欢迎的电子邮件提供商。可能想尝试NOT LIKE '%_@%_.__%'（在后面加上一个字符@）
— jonaz '16
20

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，您可以使用函数查找所有行：

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

如果您不满意在数据库中创建函数，则可以在查询中直接使用LIKE子句：

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

资源

— 埃斯波
source
7

我发现此简单的T-SQL查询对于返回有效的电子邮件地址很有用

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

PATINDEX位会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。

可以相反地执行您想要的操作：

SELECT email
FROM People
WHERE NOT (email LIKE '%_@__%.__%' 
    AND PATINDEX('%[^a-z,0-9,@,.,_]%', REPLACE(email, '-', 'a')) = 0)
— 杰克·艾伦
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2
您可以通过转义连字符来消除REPLACE功能，如下所示：AND PATINDEX('%[^a-z,0-9,@,.,_,\-]%', email) = 0
— Splendor
4

的MySQL

SELECT * FROM `emails` WHERE `email`
NOT REGEXP '[-a-z0-9~!$%^&*_=+}{\\\'?]+(\\.[-a-z0-9~!$%^&*_=+}{\\\'?]+)*@([a-z0-9_][-a-z0-9_]*(\\.[-a-z0-9_]+)*\\.(aero|arpa|biz|com|coop|edu|gov|info|int|mil|museum|name|net|org|pro|travel|mobi|[a-z][a-z])|([0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}))(:[0-9]{1,5})?'
— Zemistr
source
3
这在MS SQL Server上不起作用。请指定上述语法在其上有效的数据库供应商。
— Neolisk
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
;
— 马克斯
source
@Manishm尝试通过电子邮件发送PostgreSQL版本myname@gmail.。这就是为什么我拒绝投票的原因-过于复杂但无法正常工作。
— 1ac0 2014年
hello+world@gmail.com是一个有效的邮件地址
— 努里Tasdemir
请谨慎使用域名，因为某人@ domain.co.nz有效（请注意“ .co”部分）
— kurdtpage
.-_是有效的起始字符
— gliljas
2

在SQL Server 2016及更高版本上

CREATE FUNCTION [DBO].[F_IsEmail] (
 @EmailAddr varchar(360) -- Email address to check
)   RETURNS BIT -- 1 if @EmailAddr is a valid email address

AS BEGIN
DECLARE @AlphabetPlus VARCHAR(255)
      , @Max INT -- Length of the address
      , @Pos INT -- Position in @EmailAddr
      , @OK BIT  -- Is @EmailAddr OK
-- Check basic conditions
IF @EmailAddr IS NULL 
   OR @EmailAddr NOT LIKE '[0-9a-zA-Z]%@__%.__%' 
   OR @EmailAddr LIKE '%@%@%' 
   OR @EmailAddr LIKE '%..%' 
   OR @EmailAddr LIKE '%.@' 
   OR @EmailAddr LIKE '%@.' 
   OR @EmailAddr LIKE '%@%.-%' 
   OR @EmailAddr LIKE '%@%-.%' 
   OR @EmailAddr LIKE '%@-%' 
   OR CHARINDEX(' ',LTRIM(RTRIM(@EmailAddr))) > 0
       RETURN(0)



declare @AfterLastDot varchar(360);
declare @AfterArobase varchar(360);
declare @BeforeArobase varchar(360);
declare @HasDomainTooLong bit=0;

--Control des longueurs et autres incoherence
set @AfterLastDot=REVERSE(SUBSTRING(REVERSE(@EmailAddr),0,CHARINDEX('.',REVERSE(@EmailAddr))));
if  len(@AfterLastDot) not between 2 and 17
RETURN(0);

set @AfterArobase=REVERSE(SUBSTRING(REVERSE(@EmailAddr),0,CHARINDEX('@',REVERSE(@EmailAddr))));
if len(@AfterArobase) not between 2 and 255
RETURN(0);

select top 1 @BeforeArobase=value from  string_split(@EmailAddr, '@');
if len(@AfterArobase) not between 2 and 255
RETURN(0);

--Controle sous-domain pas plus grand que 63
select top 1 @HasDomainTooLong=1 from string_split(@AfterArobase, '.') where LEN(value)>63
if @HasDomainTooLong=1
return(0);

--Control de la partie locale en detail
SELECT @AlphabetPlus = 'abcdefghijklmnopqrstuvwxyz01234567890!#$%&‘*+-/=?^_`.{|}~'
     , @Max = LEN(@BeforeArobase)
     , @Pos = 0
     , @OK = 1


WHILE @Pos < @Max AND @OK = 1 BEGIN
    SET @Pos = @Pos + 1
    IF @AlphabetPlus NOT LIKE '%' + SUBSTRING(@BeforeArobase, @Pos, 1) + '%' 
        SET @OK = 0
END

if @OK=0
RETURN(0);

--Control de la partie domaine en detail
SELECT @AlphabetPlus = 'abcdefghijklmnopqrstuvwxyz01234567890-.'
     , @Max = LEN(@AfterArobase)
     , @Pos = 0
     , @OK = 1

WHILE @Pos < @Max AND @OK = 1 BEGIN
    SET @Pos = @Pos + 1
    IF @AlphabetPlus NOT LIKE '%' + SUBSTRING(@AfterArobase, @Pos, 1) + '%' 
        SET @OK = 0
END

if @OK=0
RETURN(0);







return(1);



END
— Esperento57
source
1

我发现这种方法更直观：

CREATE FUNCTION [dbo].[ContainsVailidEmail] (@Input varchar(250))
RETURNS bit
AS
BEGIN
  RETURN CASE
    WHEN @Input LIKE '%_@__%.__%' THEN 1
    ELSE 0
  END
END

我用以下方法称呼它：

SELECT [dbo].[ContainsVailidEmail] (Email) FROM [dbo].[User]

要么

如果只使用一次，那么为什么不将其用作具有以下规范的“计算列”：

(case when [Email] like '%_@__%.__%' then (1) else (0) end)

然后，您可以使用它而无需调用函数。

— 卡尼码
source
0

我建议我的职能：

CREATE FUNCTION [REC].[F_IsEmail] (
 @EmailAddr varchar(360) -- Email address to check
)   RETURNS BIT -- 1 if @EmailAddr is a valid email address

AS BEGIN
DECLARE @AlphabetPlus VARCHAR(255)
      , @Max INT -- Length of the address
      , @Pos INT -- Position in @EmailAddr
      , @OK BIT  -- Is @EmailAddr OK
-- Check basic conditions
IF @EmailAddr IS NULL 
   OR @EmailAddr NOT LIKE '[0-9a-zA-Z]%@__%.__%' 
   OR @EmailAddr LIKE '%@%@%' 
   OR @EmailAddr LIKE '%..%' 
   OR @EmailAddr LIKE '%.@' 
   OR @EmailAddr LIKE '%@.' 
   OR @EmailAddr LIKE '%@%.-%' 
   OR @EmailAddr LIKE '%@%-.%' 
   OR @EmailAddr LIKE '%@-%' 
   OR CHARINDEX(' ',LTRIM(RTRIM(@EmailAddr))) > 0
       RETURN(0)



declare @AfterLastDot varchar(360);
declare @AfterArobase varchar(360);
declare @BeforeArobase varchar(360);
declare @HasDomainTooLong bit=0;

--Control des longueurs et autres incoherence
set @AfterLastDot=REVERSE(SUBSTRING(REVERSE(@EmailAddr),0,CHARINDEX('.',REVERSE(@EmailAddr))));
if  len(@AfterLastDot) not between 2 and 17
RETURN(0);

set @AfterArobase=REVERSE(SUBSTRING(REVERSE(@EmailAddr),0,CHARINDEX('@',REVERSE(@EmailAddr))));
if len(@AfterArobase) not between 2 and 255
RETURN(0);

select top 1 @BeforeArobase=value from  string_split(@EmailAddr, '@');
if len(@AfterArobase) not between 2 and 255
RETURN(0);

--Controle sous-domain pas plus grand que 63
select top 1 @HasDomainTooLong=1 from string_split(@AfterArobase, '.') where LEN(value)>63
if @HasDomainTooLong=1
return(0);

--Control de la partie locale en detail
SELECT @AlphabetPlus = 'abcdefghijklmnopqrstuvwxyz01234567890!#$%&‘*+-/=?^_`.{|}~'
     , @Max = LEN(@BeforeArobase)
     , @Pos = 0
     , @OK = 1


WHILE @Pos < @Max AND @OK = 1 BEGIN
    SET @Pos = @Pos + 1
    IF @AlphabetPlus NOT LIKE '%' + SUBSTRING(@BeforeArobase, @Pos, 1) + '%' 
        SET @OK = 0
END

if @OK=0
RETURN(0);

--Control de la partie domaine en detail
SELECT @AlphabetPlus = 'abcdefghijklmnopqrstuvwxyz01234567890-.'
     , @Max = LEN(@AfterArobase)
     , @Pos = 0
     , @OK = 1

WHILE @Pos < @Max AND @OK = 1 BEGIN
    SET @Pos = @Pos + 1
    IF @AlphabetPlus NOT LIKE '%' + SUBSTRING(@AfterArobase, @Pos, 1) + '%' 
        SET @OK = 0
END

if @OK=0
RETURN(0);

return(1);

END
— Esperento57
source
0 
sel 'unismankur@yahoo#.co.in' as Email, 
case 
    when Email not like  '%@xx%' 
    AND  Email like  '%@%' 
    AND  CHAR_LENGTH(
     oTranslate(
      trim( Email),
      '._-@0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ',
      '')
     ) = 0
     then 'N' else 'Y'  end as Invalid_Email_Ind;

这对我来说很好。

— 库玛·安库（Kumar Ankur）
source
0 
SELECT Email FROM Employee WHERE NOT REGEXP_LIKE(Email, ‘[A-Z0-9._%+-]+@[A-Z0-9.-]+\.[A-Z]{2,4}’, ‘i’);
— 布莱克·迪瓦恩
source
1
您可以添加有关答案的解释以及这样做的原因。仅提供代码答案有点令人困惑。
— mnestorov
-1 
select * from users 
WHERE NOT
(     CHARINDEX(' ',LTRIM(RTRIM([Email]))) = 0 
AND  LEFT(LTRIM([Email]),1) <> '@' 
AND  RIGHT(RTRIM([Email]),1) <> '.' 
AND  CHARINDEX('.',[Email],CHARINDEX('@',[Email])) - CHARINDEX('@',[Email]) > 1 
AND  LEN(LTRIM(RTRIM([Email]))) - LEN(REPLACE(LTRIM(RTRIM([Email])),'@','')) = 1 
AND  CHARINDEX('.',REVERSE(LTRIM(RTRIM([Email])))) >= 3 
AND  (CHARINDEX('.@',[Email]) = 0 AND CHARINDEX('..',[Email]) = 0)
— 安东尼·斯塔布斯
source
-1 
go

create proc GetEmail

@name varchar(22),
@gmail varchar(22)

as

begin

declare @a varchar(22)

set select @a=substring(@gmail,charindex('@',@gmail),len(@gmail)-charindex('@',@gmail)+1)

if (@a = 'gmail.com)

insert into table_name values(@name,@gmail)

else

print 'please enter valid email address'

end
— 芬卡·雷迪
source
-2

我知道该帖子很旧，但经过3个月的时间，我遇到了各种电子邮件组合，能够使此sql用于验证电子邮件ID。

CREATE FUNCTION [dbo].[isValidEmailFormat]
(
    @EmailAddress varchar(500)
)
RETURNS bit
AS
BEGIN
    DECLARE @Result bit

    SET @EmailAddress = LTRIM(RTRIM(@EmailAddress));
    SELECT @Result =
    CASE WHEN
    CHARINDEX(' ',LTRIM(RTRIM(@EmailAddress))) = 0
    AND LEFT(LTRIM(@EmailAddress),1) <> '@'
    AND RIGHT(RTRIM(@EmailAddress),1) <> '.'
    AND LEFT(LTRIM(@EmailAddress),1) <> '-'
    AND CHARINDEX('.',@EmailAddress,CHARINDEX('@',@EmailAddress)) - CHARINDEX('@',@EmailAddress) > 2    
    AND LEN(LTRIM(RTRIM(@EmailAddress))) - LEN(REPLACE(LTRIM(RTRIM(@EmailAddress)),'@','')) = 1
    AND CHARINDEX('.',REVERSE(LTRIM(RTRIM(@EmailAddress)))) >= 3
    AND (CHARINDEX('.@',@EmailAddress) = 0 AND CHARINDEX('..',@EmailAddress) = 0)
    AND (CHARINDEX('-@',@EmailAddress) = 0 AND CHARINDEX('..',@EmailAddress) = 0)
    AND (CHARINDEX('_@',@EmailAddress) = 0 AND CHARINDEX('..',@EmailAddress) = 0)
    AND ISNUMERIC(SUBSTRING(@EmailAddress, 1, 1)) = 0
    AND CHARINDEX(',', @EmailAddress) = 0
    AND CHARINDEX('!', @EmailAddress) = 0
    AND CHARINDEX('-.', @EmailAddress)=0
    AND CHARINDEX('%', @EmailAddress)=0
    AND CHARINDEX('#', @EmailAddress)=0
    AND CHARINDEX('$', @EmailAddress)=0
    AND CHARINDEX('&', @EmailAddress)=0
    AND CHARINDEX('^', @EmailAddress)=0
    AND CHARINDEX('''', @EmailAddress)=0
    AND CHARINDEX('\', @EmailAddress)=0
    AND CHARINDEX('/', @EmailAddress)=0
    AND CHARINDEX('*', @EmailAddress)=0
    AND CHARINDEX('+', @EmailAddress)=0
    AND CHARINDEX('(', @EmailAddress)=0
    AND CHARINDEX(')', @EmailAddress)=0
    AND CHARINDEX('[', @EmailAddress)=0
    AND CHARINDEX(']', @EmailAddress)=0
    AND CHARINDEX('{', @EmailAddress)=0
    AND CHARINDEX('}', @EmailAddress)=0
    AND CHARINDEX('?', @EmailAddress)=0
    AND CHARINDEX('<', @EmailAddress)=0
    AND CHARINDEX('>', @EmailAddress)=0
    AND CHARINDEX('=', @EmailAddress)=0
    AND CHARINDEX('~', @EmailAddress)=0
    AND CHARINDEX('`', @EmailAddress)=0 
    AND CHARINDEX('.', SUBSTRING(@EmailAddress, CHARINDEX('@', @EmailAddress)+1, 2))=0
    AND CHARINDEX('.', SUBSTRING(@EmailAddress, CHARINDEX('@', @EmailAddress)-1, 2))=0
    AND LEN(SUBSTRING(@EmailAddress, 0, CHARINDEX('@', @EmailAddress)))>1
    AND CHARINDEX('.', REVERSE(@EmailAddress)) > 2
    AND CHARINDEX('.', REVERSE(@EmailAddress)) < 5  
    THEN 1 ELSE  0 END


    RETURN @Result
END

任何建议都欢迎！

— anurag_it
source
-9 
DELETE 
FROM `contatti` 
WHERE `EMail` NOT LIKE "%.it" 
  AND `EMail` NOT LIKE "%.com" 
  AND `EMail` NOT LIKE "%.fr"  
  AND `EMail` NOT LIKE "%.net"  
  AND `EMail` NOT LIKE "%.ru"  
  AND `EMail` NOT LIKE "%.eu"  
  AND `EMail` NOT LIKE "%.org"  
  AND `EMail` NOT LIKE "%.edu"  
  AND `EMail` NOT LIKE "%.uk"  
  AND `EMail` NOT LIKE "%.de"  
  AND `EMail` NOT LIKE "%.biz"  
  AND `EMail` NOT LIKE "%.ch"  
  AND `EMail` NOT LIKE "%.bg"  
  AND `EMail` NOT LIKE "%.info"  
  AND `EMail` NOT LIKE "%.br"  
  AND `EMail` NOT LIKE "%.pt"  
  AND `EMail` NOT LIKE "%.za"  
  AND `EMail` NOT LIKE "%.vn"  
  AND `EMail` NOT LIKE "%.es"  
  AND `EMail` NOT LIKE "%.in"  
  AND `EMail` NOT LIKE "%.dk"  
  AND `EMail` NOT LIKE "%.ni"  
  AND `EMail` NOT LIKE "%.ar"

并把你想要的所有扩展名

— 用户2199272
source
请编辑您的答案并设置代码格式以使其可读。
— kleopatra 2013年
1
大。这将删除来自奥地利，列支敦士登或其他有效TLD的所有有效电子邮件地址。（不能投票，没有足够的声誉）。
— urbanhusky
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