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！
答案 0 :(得分：138)
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。
几乎不可能在代码中验证电子邮件地址。
编辑：相关问题
答案 1 :(得分：19)
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找到所有行：
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 '_%@__%.__%'
答案 2 :(得分：6)
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
1.8PATINDEX位删除所有包含不在允许的az，0-9，＆＃39; @＆＃39;，＆＃39;。＆＃39;，＆＃39;中的字符的电子邮件地址。 _＆＃39; ＆安培; ＆＃39; - ＆＃39;一组字符。
可以按照你想要的方式进行反转：
SELECT email
FROM People
WHERE email LIKE '%_@__%.__%'
AND PATINDEX('%[^a-z,0-9,@,.,_]%', REPLACE(email, '-', 'a')) = 0
答案 3 :(得分：3)
<强>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})?'
答案 4 :(得分：2)
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。不得不申请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
;
答案 5 :(得分：1)
在sql server 2016及以上
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
答案 6 :(得分：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
我使用以下方法调用它：
SELECT [dbo].[ContainsVailidEmail] (Email) FROM [dbo].[User]
OR
如果您只打算使用一次，那么为什么不将它作为计算列，具有以下规范：
(case when [Email] like '%_@__%.__%' then (1) else (0) end)
然后你就可以使用它而无需调用函数。
答案 7 :(得分：0)
我提议我的功能：
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
答案 8 :(得分：0)
SELECT EmailAddress AS ValidEmail
FROM Contacts
WHERE EmailAddress LIKE '%_@__%.__%'
AND PATINDEX('%[^a-z,0-9,@,.,_,\-]%', EmailAddress) = 0
GO
答案 9 :(得分：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。
答案 10 :(得分：0)
SELECT Email FROM Employee WHERE NOT REGEXP_LIKE(Email, ‘[A-Z0-9._%+-]+@[A-Z0-9.-]+\.[A-Z]{2,4}’, ‘i’);
答案 11 :(得分：-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)
答案 12 :(得分：-1)
select *
from MailList.dbo.tblMailID
where
patindex ('%[ &'',":;!+=\/()<>]%', mailid) > 0 -- Invalid characters
or patindex ('[@.-_]%', mailid) > 0 -- Valid but cannot be starting character
or patindex ('%[@.-_]', mailid) > 0 -- Valid but cannot be ending character
or mid not like '%@%.%' -- Must contain at least one @ and one .
or mid like '%..%' -- Cannot have two periods in a row
or mid like '%@%@%' -- Cannot have two @ anywhere
or mid like '%.@%' or mailid like '%@.%' -- Cannot have @ and . next to each other
or mid like '%.cm' or mailid like '%.co' -- Camaroon or Colombia? Unlikely. Probably typos
or mid like '%.or' or mailid like '%.ne' -- Missing last letter
答案 13 :(得分：-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
答案 14 :(得分：-2)
我知道该帖子已经过时了，但经过3个月的时间和我遇到的各种电子邮件组合，能够制作这个用于验证电子邮件ID的SQL。
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
欢迎任何建议！
答案 15 :(得分：-8)
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"
并输入您想要的所有扩展程序