Question

Split the string into two strings using SUBSTRING function in SQL Server

Asked by: milani_lucie

Hi,

I have the following string:

steve@ibm.com

I want to split this string by the delimter '@' so i should get two strings

steve
ibm.com

Can you please provide me the code for this ? Please note that the email address length may be vary depending upon the user name. So no hard coding lengths please !

Thanks

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Asked On
2009-11-06 at 07:09:01ID24878043
Topics

SQL Server 2008

,

MS SQL Server

,

SQL Server 2005

Participating Experts
6
Points
500
Comments
12

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Answers

 

by: sstadPosted on 2009-11-06 at 07:15:25ID: 25759789

You can use the folliwng code 

<?php
$str = 'steve@ibm.com';
 
$parts = explode('@', $str);
 
echo $part[0] . "<br/>";
echo $part[1] . "<br/>";
 
?> 

                                              
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by: laneduncanPosted on 2009-11-06 at 07:19:00ID: 25759823

It should be something like this:

To get the first part:
username=substring (emailaddress,1,charindex('@',emailaddress)-1)

And the piece after the @ sign:
domainname=substring (emailaddress,charindex('@',emailaddress)+1,len(emailaddress))

This is off the top of my head; I'll run a quick test when I have a moment.

Is that enough to get you going?

 

by: sstadPosted on 2009-11-06 at 07:19:07ID: 25759824

Sorry made a typo. It should be the code below.

You can also limit the amount of parts you want to have by adding another parameter to the explode() function.
If you only want to pieces of text you can use the following: explode('@', '$yourstring', 2)

<?php
$str = 'steve@ibm.com';
 
$parts = explode('@', $str);
 
echo $parts[0] . "<br/>";
echo $parts[1] . "<br/>";
 
?> 
                                              
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by: matthewspatrickPosted on 2009-11-06 at 07:19:12ID: 25759825

SELECT email, LEFT(email, CHARINDEX('@', email) - 1) AS Part1, SUBSTRING(email, CHARINDEX('@', email) + 1, 100) AS Part2
FROM SomeTable

 

by: andycroftsPosted on 2009-11-06 at 07:19:45ID: 25759828

declare @email varchar(100)
set @email = 'steve@ibm.com'
select substring(@email,1,patindex('%@%',@email)-1 ) as string1,
            substring(@email,patindex('%@%',@email)+1,len(@email) ) as string2
select substring('steve@ibm.com',1,patindex('@') as string1




~

 

by: andycroftsPosted on 2009-11-06 at 07:20:06ID: 25759830

declare @email varchar(100)
set @email = 'steve@ibm.com'
select substring(@email,1,patindex('%@%',@email)-1 ) as string1,
            substring(@email,patindex('%@%',@email)+1,len(@email) ) as string2

 

by: prabhakaranbkPosted on 2009-11-06 at 07:20:22ID: 25759833

Please try the below,

Declare @email varchar(100)
Select @email='steve@ibm.com'
Select left(@email,charindex('@',@email,1)-1),Substring(@email,charindex('@',@email,1)+1,20)

                                              
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by: matthewspatrickPosted on 2009-11-06 at 07:20:30ID: 25759834

Wow, four nearly identical comments in the same minute :)

 

by: aneeshattingalPosted on 2009-11-06 at 07:44:20ID: 25760054

Select left(email,charindex('',email,1)-1),
       RIGHT(email, LEN(Email) - charindex('',email,1))

 

by: matthewspatrickPosted on 2009-11-09 at 11:52:57ID: 25779324

Mods,

I cannot comment on the PHP suggestions, but if the Asker wanted a SQL solution, then I recommend a split between:

http:#25759823 (laneduncan)
http:#25759825 (matthewspatrick)
http:#25759830 (andycrofts)
http:#25759833 (prabhakaranbk)

All four return the same result, and were posted within a few seconds of each other.

The accepted answer http:#25760054 came 25 minutes after the first correct answer, and contains a syntax error to boot.  (The CHARINDEX expression should be looking for '@', not '').

Cheers,

Patrick

To test the various suggestions, I ran:
 
DECLARE @email varchar(100)
SET @email = 'steve@ibm.com'
 
SELECT 'laneduncan' AS Expert, substring (@email,1,charindex('@',@email)-1) AS username, substring (@email,charindex('@',@email)+1,len(@email)) AS domain
union
select 'matthewspatrick', LEFT(@email, CHARINDEX('@', @email) - 1) AS Part1, SUBSTRING(@email, CHARINDEX('@', @email) + 1, 100)
union
select 'andycrofts',substring(@email,1,patindex('%@%',@email)-1 ),substring(@email,patindex('%@%',@email)+1,len(@email) )
union
Select 'prabhakaranbk',left(@email,charindex('@',@email,1)-1),Substring(@email,charindex('@',@email,1)+1,20)
 
it returns:
 
andycrofts	steve	ibm.com
laneduncan	steve	ibm.com
matthewspatrick	steve	ibm.com
prabhakaranbk	steve	ibm.com
 
when you try to run aneesh's code:
 
Select 'aneeshattingal',left(@email,charindex('',@email,1)-1)--,RIGHT(@email, LEN(@email) - charindex('',@email,1)) 
 
you get:
 
Msg 536, Level 16, State 5, Line 4
Invalid length parameter passed to the SUBSTRING function.
                                              
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by: milani_luciePosted on 2009-11-09 at 13:27:09ID: 25780281

Hi Guys,

Here is the correct solution:

SET @FirstPart = SUBSTRING(@TempLoginID, 1, CHARINDEX('@', @TempLoginID) - 1)
SET @LastPart = SUBSTRING(@TempLoginID, CHARINDEX('@', @TempLoginID) + 1, LEN(@TempLoginID) - (LEN(@FirstPart) + 1))

Thanks

20120131-EE-VQP-002

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