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Script SQL pour trouver des adresses e-mail non valides

Une importation de données a été effectuée à partir d'une base de données d'accès et aucune validation n'a été effectuée sur le champ d'adresse de messagerie. Quelqu'un at-il un script SQL capable de renvoyer une liste d'adresses électroniques non valides (@ manquant, etc.)?.

Merci!

sqlvalidationemail
55
campo
SELECT * FROM people WHERE email NOT LIKE '%_@__%.__%'

Tout ce qui est plus complexe renverra probablement de faux négatifs et sera plus lent.

La validation des adresses e-mail dans le code est pratiquement impossible.

EDIT: Questions connexes

133
Tomalak

Voici une solution rapide et facile:

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

Ensuite, vous pouvez trouver toutes les lignes en utilisant la fonction:

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

Si vous n'êtes pas satisfait de créer une fonction dans votre base de données, vous pouvez utiliser la clause LIKE directement dans votre requête:

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

La source

19
Espo

Je trouve cette requête T-SQL simple utile pour renvoyer des adresses électroniques valides

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

Le bit PATINDEX élimine toutes les adresses de messagerie contenant des caractères qui ne figurent pas dans les ensembles de caractères autorisés a-z, 0-9, '@', '.', '_' & '-'.

Cela peut être inversé pour faire ce que vous voulez comme ceci:

SELECT email
FROM People
WHERE NOT (email LIKE '%_@__%.__%' 
    AND PATINDEX('%[^a-z,0-9,@,.,_]%', REPLACE(email, '-', 'a')) = 0)
5
Jack Allan

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})?'
3
Zemistr
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

Cela a fonctionné pour moi. Dû appliquer rtrim et ltrim pour éviter les faux positifs. 

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

Version Postgres:

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
;
2
Manishm

Je trouve cette approche plus intuitive:

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

Je l'appelle en utilisant ce qui suit:

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

OR

Si vous n'utilisez cette option qu'une seule fois, pourquoi ne pas l'utiliser comme colonne calculée, avec la spécification suivante:

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

Ensuite, vous pouvez simplement l'utiliser sans avoir à appeler une fonction.

1
CarneyCode

Sur le serveur SQL 2016 et plus

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
1
Esperento57
SELECT EmailAddress AS ValidEmail
FROM Contacts
WHERE EmailAddress LIKE '%_@__%.__%'
        AND PATINDEX('%[^a-z,0-9,@,.,_,\-]%', EmailAddress) = 0
GO

Veuillez vérifier ce lien: https://blog.sqlauthority.com/2017/11/12/validate-email-address-sql-server-interview-question-week-147/

0
rino
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;

Cela fonctionne très bien pour moi.

0
Kumar Ankur

Je propose ma fonction:

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
0
Esperento57