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Trim Everything After @ In Email Address

Posted on 2003-12-03
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Last Modified: 2008-03-17
I am trying to write a statement that will take everything in an email address from the @ and all characters to its right off of the returned value.

i.e. - thisisatest@testing123456789.com  in this case the only data that will be returned would be

        thisisatest
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Question by:H-SC
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Expert Comment

by:TimCottee
ID: 9866573
Hi H-SC,

Select Left(EMailAddress,CharIndex('@',EMailAddress) -1) As Address From MyTable

Is one way, this finds the position of the @ and then returns the left portion of the field.

Tim Cottee MCSD, MCDBA, CPIM
Brainbench MVP for Visual Basic
http://www.brainbench.com
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by:H-SC
ID: 9866653
Thanks for the quick response.

When I run that statement it gives me an error "Invalid param passed to the substring function"
error #536
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Expert Comment

by:TimCottee
ID: 9866694
The most likely situation is that one or more of the entries either contain no address or the address does not have the @ character in it. There are a number of ways to work around this, perhaps the easiest is to append an @ to the string in the first place which means you will get the entire address if no @ already present or the correct information if one is.

Select Left(EMailAddress + '@',CharIndex('@',EMailAddress + '@') -1) As Address From MyTable
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by:H-SC
ID: 9866747
That is great.  I do not get errors on the last statement, however it only returns the first character in the email address.

i.e. - in my example of thisisatest@testing123456789.com would return "t"

any ideas?
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Expert Comment

by:TimCottee
ID: 9866782
I am using this just to prove the case:

declare @EmailAddress varchar(50)
Set @emailaddress = 'thisisatest@testing123456789.com'
Select Left(@EMailAddress + '@',CharIndex('@',@EMailAddress + '@') -1) As Address

Which is returning the correct value.
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Author Comment

by:H-SC
ID: 9866860
I ran the last one and it works.  Will the declare or set change any data or write back?
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Expert Comment

by:TimCottee
ID: 9866900
No not at all, this is unconnected to any table and just to demonstrate the principal. If the equivalent statement using your table and its fieldname(s) still fails then there must be something different with the data in the column that is causing this problem.
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Author Comment

by:H-SC
ID: 9866935
GREAT!  will I need to do a different statement that will return real-time values.  The last one returned one row; "thisisatest"
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