Question

我的表中的数据如下所示：

select email_body from email_table


email_body
----------
Ashely, call me.   thanks  --- Original message ---  From: Ashley  To: Lee Subject: Homework  Sent: 3/6/2017 2:32:54 AM GMT  I have a quick question.
Mike, I have all the data you need.  Jim   --- Original message ---  From: Mike  To: Jim Subject: Not Sure  Sent: 3/18/2017 12:02:51 AM GMT  Are you available to go over this?
William, Are you around.   thanks  --- Original message ---  From: Joe  To: William Subject: Nothing much  Sent: 4/16/2017 4:17:23 PM GMT  I need some sleep.
Joan, call me.   Ralph  --- Original message ---  From: Ralph  To: Joan Subject: I need help  Sent: 3/30/2017 5:12:50 AM GMT  Call Rich.

我想返回email_body中列出的日期和时间：

Results:

Original_message
----------------
3/6/2017 2:32:54 AM
3/18/2017 12:02:51 AM
4/16/2017 4:17:23 PM
3/30/2017 5:12:50 AM

Answer 1

因为日期长度可能会有所不同（例如1/1/2017与12/12/2017），你需要找到它的起始位置（很容易 - Sent:之后的6个字符）和它结束的位置（在M GMT之前是2个字符）。剩下的工作是substring完成的。

SELECT SUBSTRING(email_body, CHARINDEX('Sent:', email_body) + 6,
 CHARINDEX('M GMT ',email_body) - 2 - CHARINDEX('Sent:', email_body) - 6) as Original_message
FROM email_table

Answer 2

您可以使用类似

的内容

select SUBSTRING (email_body , PATINDEX ( 'Subject' ,email_body )+1, 18) from email_table

Answer 3

如果您对Table-Valued-Function开放，它将根据模式 AND 快速安全地提取一个或多个值，您会厌倦所有必需的字符串操作，请考虑以下内容：

示例

Declare @email_table table (id int,email_body varchar(max)) Insert Into @email_table values (1,'Ashely, call me. thanks --- Original message --- From: Ashley To: Lee Subject: Homework Sent: 3/6/2017 2:32:54 AM GMT I have a quick question.'), (2,'Mike, I have all the data you need. Jim --- Original message --- From: Mike To: Jim Subject: Not Sure Sent: 3/18/2017 12:02:51 AM GMT Are you available to go over this?'), (3,'William, Are you around. thanks --- Original message --- From: Joe To: William Subject: Nothing much Sent: 4/16/2017 4:17:23 PM GMT I need some sleep.'), (4,'Joan, call me. Ralph --- Original message --- From: Ralph To: Joan Subject: I need help which was Sent: 3/30/2017 5:12:50 AM GMT Call Rich.') Select A.ID ,Original_message = B.RetVal From @email_table A Cross Apply [dbo].[udf-Str-Extract](A.email_body,' Sent: ',' GMT ') B

<强>返回

ID Original_message 1 3/6/2017 2:32:54 AM 2 3/18/2017 12:02:51 AM 3 4/16/2017 4:17:23 PM 4 3/30/2017 5:12:50 AM

实际的TVF返回

RetSeq RetPos RetLen RetVal 1 101 19 3/6/2017 2:32:54 AM 1 115 21 3/18/2017 12:02:51 AM 1 114 20 4/16/2017 4:17:23 PM 1 111 20 3/30/2017 5:12:50 AM

感兴趣的UDF

CREATE FUNCTION [dbo].[udf-Str-Extract] (@String varchar(max),@Delimiter1 varchar(100),@Delimiter2 varchar(100)) Returns Table As Return ( with cte1(N) As (Select 1 From (Values(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) N(N)), cte2(N) As (Select Top (IsNull(DataLength(@String),0)) Row_Number() over (Order By (Select NULL)) From (Select N=1 From cte1 N1,cte1 N2,cte1 N3,cte1 N4,cte1 N5,cte1 N6) A ), cte3(N) As (Select 1 Union All Select t.N+DataLength(@Delimiter1) From cte2 t Where Substring(@String,t.N,DataLength(@Delimiter1)) = @Delimiter1), cte4(N,L) As (Select S.N,IsNull(NullIf(CharIndex(@Delimiter1,@String,s.N),0)-S.N,8000) From cte3 S) Select RetSeq = Row_Number() over (Order By N) ,RetPos = N ,RetLen = charindex(@Delimiter2,RetVal)-1 ,RetVal = left(RetVal,charindex(@Delimiter2,RetVal)-1) From (Select A.N,RetVal = ltrim(rtrim(Substring(@String, A.N, A.L))) From cte4 A ) A Where charindex(@Delimiter2,RetVal)>1 ) /* Max Length of String 1MM characters Declare @String varchar(max) = 'Dear [[FirstName]] [[LastName]], ...' Select * From [dbo].[udf-Str-Extract] (@String,'[[',']]') */

从列中选择特定字符串

3 个答案: