使用 tsql 从字符串中提取电子邮件地址

Question

我正在尝试从现有 cmets 字段中提取电子邮件地址并将其放入自己的列中。字符串可能是这样的“this is an example comment with an email address of someemail@domain.org”，或者只是电子邮件本身“someemail@domain.org”。

我认为最好的办法是找到'@'符号的索引并在两个方向上搜索，直到字符串的末尾被命中或有空格。谁能帮我完成这个实现？

Answer 1

我知道 wewestthemenace 已经回答了这个问题，但他/她的解决方案似乎过于复杂。为什么要将电子邮件地址的左侧和右侧连接在一起？我宁愿只找到电子邮件地址的开头和结尾，然后使用子字符串返回电子邮件地址，如下所示：

我的桌子

DECLARE @Table TABLE (comment NVARCHAR(50));
INSERT INTO @Table
VALUES ('blah MyEmailAddress@domain.org'),            --At the end
        ('blah MyEmailAddress@domain.org blah blah'), --In the middle
        ('MyEmailAddress@domain.org blah'),           --At the beginning
        ('no email');

实际查询：

SELECT  comment,        
        CASE
            WHEN CHARINDEX('@',comment) = 0 THEN NULL
            ELSE SUBSTRING(comment,beginningOfEmail,endOfEmail-beginningOfEmail)
        END email
FROM @Table
CROSS APPLY (SELECT CHARINDEX(' ',comment + ' ',CHARINDEX('@',comment))) AS A(endOfEmail)
CROSS APPLY (SELECT DATALENGTH(comment)/2 - CHARINDEX(' ',REVERSE(' ' + comment),CHARINDEX('@',REVERSE(' ' + comment))) + 2) AS B(beginningOfEmail)

结果：

comment                                            email
-------------------------------------------------- --------------------------------------------------
blah MyEmailAddress@domain.org                     MyEmailAddress@domain.org
blah MyEmailAddress@domain.org blah blah           MyEmailAddress@domain.org
MyEmailAddress@domain.org blah                     MyEmailAddress@domain.org
no email                                           NULL

Answer 2

您可以在字符串中搜索'@'。然后在'@' 的LEFT 和RIGHT 一侧获得字符串。然后你想REVERSELEFT 一侧并获得' ' 的第一次出现，然后从那里获得SUBSTRING。然后REVERSE就可以得到原始表格了。相同的原则适用于RIGHT 侧，无需执行REVERSE。

示例字符串：'some text someemail@domain.org some text'

1. LEFT = '一些文本 someemail'
2. RIGHT = '@domain.org 一些文字'
3. 反向 LEFT = 'liameemos txet emos'
4. SUBSTRING 直到第一个空格 = 'liameemos'
5. REVERSE(4) = 一些电子邮件
6. SUBSTRING (2) 直到第一个空格 = '@domain.org'
7. 组合 5 和 6 = 'someemail@domain.org'

您的查询将是：

;WITH CteEmail(email) AS(
    SELECT 'someemail@domain.org' UNION ALL
    SELECT 'some text someemail@domain.org some text' UNION ALL
    SELECT 'no email'
)
,CteStrings AS(
    SELECT
        [Left] = LEFT(email, CHARINDEX('@', email, 0) - 1),
        Reverse_Left = REVERSE(LEFT(email, CHARINDEX('@', email, 0) - 1)),
        [Right] = RIGHT(email, CHARINDEX('@', email, 0) + 1)
    FROM CteEmail
    WHERE email LIKE '%@%'
)
SELECT *,
    REVERSE(
        SUBSTRING(Reverse_Left, 0, 
            CASE
                WHEN CHARINDEX(' ', Reverse_Left, 0) = 0 THEN LEN(Reverse_Left) + 1
                ELSE CHARINDEX(' ', Reverse_Left, 0)
            END
        )
    )
    +
    SUBSTRING([Right], 0,
        CASE
            WHEN CHARINDEX(' ', [Right], 0) = 0 THEN LEN([Right]) + 1
            ELSE CHARINDEX(' ', [Right], 0)
        END
    )
FROM CteStrings

样本数据：

email
----------------------------------------
someemail@domain.org
some text someemail@domain.org some text
no email

结果

---------------------
someemail@domain.org
someemail@domain.org

Answer 3

Stephan 的答案非常适合在每一行中查找单个电子邮件地址。

但是，当我尝试在每一行中获取多个电子邮件地址时遇到了这个错误：

传递给 LEFT 或 SUBSTRING 函数的长度参数无效

我使用this answer from DBA Stack Exchange 来获取字符串中@ 的所有位置。它需要一个表值函数，该函数返回等于字符串中某个模式的数量的位置数。我还必须修改 CROSS APPLY 函数来处理多个电子邮件地址。

我的桌子：

DECLARE @Table TABLE (comment VARCHAR(500));
INSERT INTO @Table (comment)
VALUES ('blah blah My.EmailAddress@domain.org more blah someemailaddress@domain.com even more blah asdf@gmail.com'),
       ('blah hello.world@domain.org more'),
       ('no email')

表值函数：

CREATE FUNCTION dbo.fnFindPatternLocation
(
    @string NVARCHAR(MAX),
    @term   NVARCHAR(255)
)
RETURNS TABLE
AS
    RETURN 
    (
        SELECT pos = Number - LEN(@term) 
        FROM (SELECT Number, Item = LTRIM(RTRIM(SUBSTRING(@string, Number, 
        CHARINDEX(@term, @string + @term, Number) - Number)))
        FROM (SELECT ROW_NUMBER() OVER (ORDER BY [object_id])
        FROM sys.all_objects) AS n(Number)
        WHERE Number > 1 AND Number <= CONVERT(INT, LEN(@string))
        AND SUBSTRING(@term + @string, Number, LEN(@term)) = @term
    ) AS y);
GO

查询：

SELECT comment, pos, SUBSTRING(comment,beginningOfEmail,endOfEmail-beginningOfEmail) AS email
FROM @Table
CROSS APPLY (SELECT pos FROM dbo.fnFindPatternLocation(comment, '@')) AS A(pos)
CROSS APPLY (SELECT CHARINDEX(' ',comment + ' ', pos)) AS B(endOfEmail)
CROSS APPLY (SELECT pos - CHARINDEX(' ', REVERSE(SUBSTRING(comment, 1, pos))) + 2) AS C(beginningOfEmail)

结果：

comment
---------------------------------------------------------------------------------------------------------
blah blah My.EmailAddress@domain.org more blah someemailaddress@domain.com even more blah asdf@gmail.com
blah blah My.EmailAddress@domain.org more blah someemailaddress@domain.com even more blah asdf@gmail.com
blah blah My.EmailAddress@domain.org more blah someemailaddress@domain.com even more blah asdf@gmail.com
blah hello.world@domain.org more

---

pos    email
---    ------------------------------
26     My.EmailAddress@domain.org
64     someemailaddress@domain.com
95     asdf@gmail.com
17     hello.world@domain.org

Answer 4

DECLARE @t TABLE (row_id INT, email VARCHAR(100))

INSERT @t (row_id, email)
VALUES (1, 'drgkls<ivan@gvi.ru>, info@gvi.com, @ dgh507-16-65@'),
        (2, 'hjshfkjshfj@kjs.kjsehf herwfjewr@kjsd.com adjfhja@.com u3483dhj@hhb@.dfj'),
        (3, 'kjsdghfjs4254.23detygh@jhjdfg.dgb лдоврывплдоо isgfsi@ klsdfksdl@,dd.com')

DECLARE @pat VARCHAR(100) = '%[^a-z0-9@._ ]%';

WITH f AS (
         SELECT    row_id,
                 CAST(' ' + email + ' ' AS VARCHAR(102)) email,
                 SUBSTRING(email, PATINDEX(@pat, email), 1) bad,
                 PATINDEX(@pat, email) pat
         FROM    @t
         UNION ALL
         SELECT    row_id,
                 CAST(REPLACE(email, bad, ' ') AS VARCHAR(102)),
                 SUBSTRING(REPLACE(email, bad, ' '), PATINDEX(@pat, REPLACE(email, bad, ' ')), 1) bad,
                 PATINDEX(@pat, REPLACE(email, bad, ' '))
         FROM    f
         WHERE    PATINDEX(@pat, email) > 0
     ),
     s AS 
     (
         SELECT    row_id,
                 email, PATINDEX('%@%', email) pos 
         FROM    f 
         WHERE    pat = 0
                 AND    PATINDEX('%@%', email) > 0
         UNION ALL
         SELECT    row_id,
                 SUBSTRING(email, pos + 1, 102), 
                 PATINDEX('%@%', SUBSTRING(email, pos + 1, 102))
         FROM    s
         WHERE    PATINDEX('%@%', SUBSTRING(email, pos + 1, 102)) > 0
     )

SELECT  row_id, o1 + pp
FROM    s   
        CROSS APPLY (SELECT    REVERSE(LEFT(email, pos -1)) s1) x
        CROSS APPLY (SELECT    CHARINDEX(' ', s1) i1) y
        CROSS APPLY (SELECT    REVERSE(LEFT(s1, i1 -1)) o1 WHERE i1 > 0) z
        CROSS APPLY (SELECT    CHARINDEX(' ', email, pos) i2) e
        CROSS APPLY (SELECT    SUBSTRING(email, pos, i2 -pos) pp WHERE    i2 > pos + 1) q
WHERE    LEN(o1) > 1
        AND CHARINDEX('.', pp) > 0
        AND PATINDEX('%@%@%', pp) = 0
        AND PATINDEX('%@.%', pp) = 0
        AND PATINDEX('%.', pp) = 0

Answer 5

这一行也可以（虽然有点长，但哈哈）：

--declare @a varchar(100) 
--set @a = 'a asfd saasd asdfgh@asd.com wqe z zx cxzc '
select substring(substring(@a,0,charindex('@',@a)),len(substring(@a,0,charindex('@',@a)))-charindex(' ',reverse(substring(@a,0,charindex('@',@a))))+2,len(substring(@a,0,charindex('@',@a)))) + substring(substring(@a,charindex('@',@a),len(@a)),0,charindex(' ',substring(@a,charindex('@',@a),len(@a))))

Answer 6

对于包含换行符的字符串，我使用 PATINDEX 修改了 Felix 的答案，以搜索第一个控制字符而不是空格。

我还必须修改 Right 字段以减去正确的文本数量。

    WITH CteEmail(email) AS(
        SELECT 'example string with new lines

    Email: some.example@email.address.com
(first email address - should be returned)

    Email: another@test.co.uk
(other email addresses should be ignored

more example text' UNION ALL
        SELECT 'Email: some.example@email.address.com' UNION ALL
        SELECT 'someemail@domain.org' UNION ALL
        SELECT 'some text someemail@domain.org some text' UNION ALL
        SELECT 'no email'
    )
    ,CteStrings AS(
        SELECT
            [Left] = LEFT(email, CHARINDEX('@', email, 0) - 1),
            Reverse_Left = REVERSE(LEFT(email, CHARINDEX('@', email, 0) - 1)),
            [Right] = RIGHT(email, LEN(email) - CHARINDEX('@', email, 0) + 1 )
        FROM CteEmail
        WHERE email LIKE '%@%'
    )
    SELECT *,
        REVERSE(
            SUBSTRING(Reverse_Left, 0, 
                CASE
                    WHEN PATINDEX('%[' + CHAR(10)+'- ]%', Reverse_Left) = 0 THEN LEN(Reverse_Left) + 1
                    ELSE PATINDEX('%[' + CHAR(0)+'- ]%', Reverse_Left)
                END
            )
        )
        +
        SUBSTRING([Right], 0,
            CASE
                WHEN PATINDEX('%[' + CHAR(0)+'- ]%', [Right]) = 0 THEN LEN([Right]) + 1
                ELSE PATINDEX('%[' + CHAR(0)+'- ]%', [Right])
            END
        )
    FROM CteStrings

Answer 7

如果您在函数中需要它，那么这对我有用...

CREATE FUNCTION [dbo].[extractEmail]
(
    @input nvarchar(500)
)
RETURNS nvarchar(100)
AS
BEGIN
    DECLARE @atPosition int
    DECLARE @firstRelevantSpace int
    DECLARE @name nvarchar(100)
    DECLARE @secondRelelvantSpace int
    DECLARE @everythingAfterAt nvarchar(500)
    DECLARE @domain nvarchar(100)
    DECLARE @email nvarchar(100) = ''
    IF CHARINDEX('@', @input,0) > 0
    BEGIN
        SET @input = ' ' + @input
        SET @atPosition = CHARINDEX('@', @input, 0)
        SET @firstRelevantSpace = CHARINDEX(' ',REVERSE(LEFT(@input, CHARINDEX('@', @input, 0) - 1)))
        SET @name = REVERSE(LEFT(REVERSE(LEFT(@input, @atPosition - 1)),@firstRelevantSpace-1))
        SET @everythingAfterAt = SUBSTRING(@input, @atPosition,len(@input)-@atPosition+1)
        SET @secondRelelvantSpace = CHARINDEX(' ',@everythingAfterAt)
        IF @secondRelelvantSpace = 0
            SET @domain = @everythingAfterAt
        ELSE
            SET @domain = LEFT(@everythingAfterAt, @secondRelelvantSpace)
        SET @email = @name + @domain
    END
    RETURN @email
END

Answer 8

使用 Cymorg 的函数：我遇到了一个问题，我的数据包含 CR/LF，它阻止了函数 100% 工作。很难弄清楚，因为在 select 语句中使用该函数时，它偶尔会返回不正确的结果。如果我从我的查询结果中复制了有问题的文本并使用 sql print 和引号中的文本调用该函数，它会正常工作。不可思议！

经过多次尝试和错误，我使用 sql replace 将 CR/LF 替换为空格和 huzza！我是一个很好的猜测者。

select extractEmail(replace(replace(MyColumn,CHAR(10),' '),CHAR(13),' ')) as AsYouWish from FacilityContacts