Скрипт Sql для поиска недействительных адресов электронной почты

Question

Импорт данных был выполнен из базы данных доступа, и в поле адреса электронной почты не было проверки. У кого-нибудь есть сценарий sql, который может вернуть список недействительных адресов электронной почты (отсутствует @ и т. Д.).

Спасибо!

Answer 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

Answer 2

Я знаю, что сообщение старое, но через 3 месяца и с различными комбинациями электронной почты, с которыми я столкнулся, я смог сделать этот sql для проверки идентификаторов электронной почты.

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

Любые предложения приветствуются!

Answer 3

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

Answer 4

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"

и поставь все нужные расширения

Answer 5

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)