好吧，有一百万个正则表达式用于验证电子邮件地址，但是一些可以集成到 Sql Server 2005 的 TSQL 查询中的基本电子邮件验证怎么样？
我不想使用 CLR 过程或函数。 只是直接的 TSQL。
有人已经解决这个问题了吗？
非常基本是：
SELECT
EmailAddress,
CASE WHEN EmailAddress LIKE '%_@_%_.__%'
AND EmailAddress NOT LIKE '%[any obviously invalid characters]%'
THEN 'Could be'
ELSE 'Nope'
END Validates
FROM
Table
这与中间带有 @ 的所有内容相匹配，前面至少有一个字符，后跟至少两个字符、一个点和至少两个 TLD。
您可以编写更多
LIKE无论如何，在数据库级别执行此操作可能是错误的方法，因此如上所述的基本健全性检查可能是性能方面最好的选择，并且在应用程序中执行此操作将为您提供更大的灵活性。
这是一个更详细的示例函数，我不记得我从哪里得到这个（几年前），或者我是否修改过它，否则我会包含正确的归属：
CREATE FUNCTION [dbo].[fnAppEmailCheck](@email VARCHAR(255))
--Returns true if the string is a valid email address.
RETURNS bit
as
BEGIN
DECLARE @valid bit
IF @email IS NOT NULL
SET @email = LOWER(@email)
SET @valid = 0
IF @email like '[a-z,0-9,_,-]%@[a-z,0-9,_,-]%.[a-z][a-z]%'
AND LEN(@email) = LEN(dbo.fnAppStripNonEmail(@email))
AND @email NOT like '%@%@%'
AND CHARINDEX('.@',@email) = 0
AND CHARINDEX('..',@email) = 0
AND CHARINDEX(',',@email) = 0
AND RIGHT(@email,1) between 'a' AND 'z'
SET @valid=1
RETURN @valid
END
很好的答案！ 根据这些建议，我想出了一个结合了最好的 2 个答案的简化函数。
CREATE FUNCTION [dbo].[fnIsValidEmail]
(
@email varchar(255)
)
--Returns true if the string is a valid email address.
RETURNS bit
As
BEGIN
RETURN CASE WHEN ISNULL(@email, '') <> '' AND @email LIKE '%_@%_.__%' THEN 1 ELSE 0 END
END
在 SQL 2016 或 + 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!#$%&‘*+-/=?^_`.{|}~'
, @Max = LEN(@BeforeArobase)
, @Pos = 0
, @OK = 1
WHILE @Pos < @Max AND @OK = 1 BEGIN
SET @Pos = @Pos + 1
IF @AlphabetPlus NOT LIKE '%' + SUBSTRING(@BeforeArobase, @Pos, 1) + '%'
SET @OK = 0
END
if @OK=0
RETURN(0);
--Control de la partie domaine en detail
SELECT @AlphabetPlus = 'abcdefghijklmnopqrstuvwxyz01234567890-.'
, @Max = LEN(@AfterArobase)
, @Pos = 0
, @OK = 1
WHILE @Pos < @Max AND @OK = 1 BEGIN
SET @Pos = @Pos + 1
IF @AlphabetPlus NOT LIKE '%' + SUBSTRING(@AfterArobase, @Pos, 1) + '%'
SET @OK = 0
END
if @OK=0
RETURN(0);
return(1);
END
FnAppStripNonEmail 在分数下缺失，需要将其添加到保留值中
Create Function [dbo].[fnAppStripNonEmail](@Temp VarChar(1000))
Returns VarChar(1000)
AS
Begin
Declare @KeepValues as varchar(50)
Set @KeepValues = '%[^a-z,0-9,_,@,.,-]%'
While PatIndex(@KeepValues, @Temp) > 0
Set @Temp = Stuff(@Temp, PatIndex(@KeepValues, @Temp), 1, '')
Return @Temp
End
CREATE FUNCTION fnIsValidEmail
(
@email varchar(255)
)
RETURNS bit
AS
BEGIN
DECLARE @IsValidEmail bit = 0
IF (@email not like '%[^a-z,0-9,@,.,!,#,$,%%,&,'',*,+,--,/,=,?,^,_,`,{,|,},~]%' --First Carat ^ means Not these characters in the LIKE clause. The list is the valid email characters.
AND @email like '%_@_%_.[a-z,0-9][a-z]%'
AND @email NOT like '%@%@%'
AND @email NOT like '%..%'
AND @email NOT like '.%'
AND @email NOT like '%.'
AND CHARINDEX('@', @email) <= 65
)
BEGIN
SET @IsValidEmail = 1
END
RETURN @IsValidEmail
END
Create Function [dbo].[fnAppStripNonEmail](@Temp VarChar(1000))
Returns VarChar(1000)
AS
Begin
Declare @KeepValues as varchar(50)
Set @KeepValues = '%[^a-z,0-9,@,.,-]%'
While PatIndex(@KeepValues, @Temp) > 0
Set @Temp = Stuff(@Temp, PatIndex(@KeepValues, @Temp), 1, '')
Return @Temp
End
这是选择它们的最简单方法。
使用此查询
SELECT * FROM <TableName> WHERE [EMail] NOT LIKE '%_@__%.__%'
从 sql 2008 开始 patindex 删除重复查找
CREATE FUNCTION ValidateEmail (@Email VARCHAR(100))
RETURNS VARCHAR(3)
AS
BEGIN
RETURN (
SELECT CASE
WHEN ISNULL(@Email, '') = '' THEN 'No'
WHEN @Email LIKE '% %' THEN 'No'
WHEN PATINDEX('%["(),:;<>]%', @Email) <> 0 THEN 'No'
WHEN SUBSTRING(@Email, CHARINDEX('@', @Email), LEN(@Email)) LIKE ('%[!#$%&*+/=?^`_{|]%') THEN 'No'
WHEN (LEFT(@Email, 1) LIKE ('[-_.+]') OR RIGHT(@Email, 1) LIKE ('[-_.+]')) THEN 'No'
WHEN (@Email LIKE '%[%' OR @Email LIKE '%]%') THEN 'No'
WHEN @Email LIKE '%@%@%' OR @Email LIKE '%@%..%' THEN 'No'
WHEN PATINDEX('_%@_%.__%', @Email) = 0 THEN 'No'
ELSE 'Yes'
END
)
END;
虽然个人我会使用
trim(@Email) = '' THEN 'No'@Email LIKE '% %' THEN 'No'来自托马拉克的精选
select 1
where @email not like '%[^a-z,0-9,@,.]%'
and @email like '%_@_%_.__%'