Question

我有一个查询，其中一列是电子邮件标题的字符串，例如：

From: Media Temple user (mt.kb.user@gmail.com)
Subject: article: How to Trace a Email
Date: January 25, 2011 3:30:58 PM PDT
To: user@example.com
Envelope-To: user@example.com
Delivery-Date: Tue, 25 Jan 2011 15:31:01 -0700
Domainkey-Signature: a=rsa-sha1; c=nofws; d=gmail.com; s=gamma; h=message-id:da...
Message-Id: <c8f49cec0807011530k11196ad4p7cb4b9420f2ae752@mail.gmail.com>
Mime-Version: 1.0
Content-Type: multipart/alternative; boundary="----=_Part_3927_12044027.1214951...
Message Body: This is a KnowledgeBase article that provides information on how ...

我想仅提取'To:'字段中包含的电子邮件地址，在上面的示例user@example.com中。

我该如何做到这一点？

Answer 1

您可以使用分割功能。我喜欢使用数字表的版本，但是there are many alternatives。首先，一个包含1,000,000行的数字表：

SET NOCOUNT ON;
DECLARE @UpperLimit INT;
SET @UpperLimit = 1000000;

WITH n(rn) AS
(
    SELECT TOP (@UpperLimit) ROW_NUMBER() OVER (ORDER BY s1.[object_id])
    FROM sys.all_columns AS s1, sys.all_objects ORDER BY s1.[object_id]
)
SELECT [Number] = rn - 1
INTO dbo.Numbers FROM n
WHERE rn <= @UpperLimit + 1;

CREATE UNIQUE CLUSTERED INDEX n ON dbo.Numbers([Number]);

现在是一个通用的，内联的表值分割函数，它将分隔的字符串转换为集合：

CREATE FUNCTION dbo.SplitString
(
    @List NVARCHAR(MAX),
    @Delim VARCHAR(255)
)
RETURNS TABLE
AS
    RETURN ( SELECT [Value] FROM 
      ( 
        SELECT 
          [Value] = LTRIM(RTRIM(SUBSTRING(@List, [Number],
          CHARINDEX(@Delim, @List + @Delim, [Number]) - [Number])))
        FROM dbo.Numbers WHERE Number <= LEN(@List)
        AND SUBSTRING(@Delim + @List, [Number], LEN(@Delim)) = @Delim
      ) AS x
    );
GO

然后很简单：

DECLARE @x NVARCHAR(MAX) = N'From: Media Temple user (mt.kb.user@gmail.com)
Subject: article: How to Trace a Email
Date: January 25, 2011 3:30:58 PM PDT
To: user@example.com
Return-Path: <mt.kb.user@gmail.com>
Envelope-To: user@example.com
...';

SELECT LTRIM(SUBSTRING(Value, 4, 4000)) 
  FROM dbo.SplitString(@x, CHAR(13)+CHAR(10))
  WHERE Value LIKE 'To: %@%';

表中的数据？好的，没问题：

DECLARE @a TABLE(id INT, email NVARCHAR(MAX));

INSERT @a VALUES
(1,N'From: Media Temple user (mt.kb.user@gmail.com)
Subject: article: How to Trace a Email
Date: January 25, 2011 3:30:58 PM PDT
To: user@example.com
Return-Path: <mt.kb.user@gmail.com>
Envelope-To: user@example.com
...'),
(2,N'From: Media Temple user (mt.kb.user@gmail.com)
Subject: article: How to Trace a Email
Date: January 25, 2011 3:30:58 PM PDT
To: differentUser@somewhereelse.com
Return-Path: <mt.kb.user@gmail.com>
Envelope-To: user@example.com
...');

SELECT a.id, LTRIM(SUBSTRING(x.Value, 4, 4000))
FROM @a AS a
CROSS APPLY dbo.SplitString(a.email, CHAR(13)+CHAR(10)) AS x
WHERE x.Value LIKE 'To: %@%';

现在，您可能需要使用分隔符 - 它可能只是CHAR（10），或者只是CHAR（13），或者它们可能处于不同的顺序 - 不确定，并且无法从您的代码中分辨出来这是什么......

Answer 2

您可以使用XML功能拆分行并找到所需内容;

DECLARE @X XML

SELECT @X = CONVERT(XML, '<y><x>' + 
                REPLACE(REPLACE(value, '<', '&lt;'), CHAR(10), '</x><x>') + 
                 '</x></y>')
FROM test

SELECT [Value] = T.c.value('.','NVARCHAR(MAX)')
FROM @X.nodes('/y/x') T(c)
WHERE T.c.value('.','NVARCHAR(MAX)') LIKE 'To: %'

An SQLfiddle to test with

Answer 3

试试这个：

select substring(@s, charindex(char(13)+char(10)+'To: ', @s) + 6, charindex(char(13), @s, charindex(char(13)+char(10)+'To: ', @s)+6) - (charindex(char(13)+char(10)+'To: ', @s)+6))

这是一个完整的测试脚本：

declare @s varchar(500)

set @s = 'Date: January 25, 2011 3:30:58 PM PDT
To: user@example.com
Return-Path: <mt.kb.user@gmail.com>
Envelope-To: user@example.com'

select substring(@s, charindex(char(13)+char(10)+'To: ', @s) + 6, charindex(char(13)+char(10), @s, charindex(char(13)+char(10)+'To: ', @s)+6) - (charindex(char(13)+char(10)+'To: ', @s)+6))

请注意，在正确的电子邮件中，标题必须根据规范RFC2822由CRLF（char（13）+ char（10））分隔，并且上述代码做出相同的假设。

如果您的电子邮件中有不同的行结尾，则可能必须将char(13)+char(10)的每次出现更改为仅char(13)或char(10)。如果您这样做，请记住同时将+6调整为+5（因为它减少了一个字符）。

Answer 4

如果电子邮件地址位于第一个'To:'和'Return-Path:'之间，您可以使用此地址（ Fiddle demo ）：

declare @s nvarchar(max) = 'From: Media Temple user (mt.kb.user@gmail.com)
                        Subject: article: How to Trace a Email
                        Date: January 25, 2011 3:30:58 PM PDT
                        To: user@example.com
                        Return-Path: <mt.kb.user@gmail.com>...'

select substring(@s, charindex('To:',@s)+3, 
             charindex('Return-Path:',@s)- charindex('To:',@s)-3)

--Results
user@example.com

更通用的版本：假设电子邮件地址位于第一个返回路径之前

;with cte as (
 select reverse(left(@s, charindex('Return-Path:',@s)-1)) rs
)
select reverse(left(rs, charindex(':oT', rs)-1)) 
from cte

在表格查询中，请将@s 替换为您的column name。

提取电子邮件标题的“收件人：”字段

4 个答案: