sql statement to get before '@' from email address

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sql statement to get before '@' from email address

scuby.me

Mr or Mrs. 500

Points: 587
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October 25, 2013 at 1:25 pm

#395726

i have email address column - pearson.charles@companyname.com
i want to show pearson.charles..anything before '@'
pls help

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p.shabbir SSCrazy Points: 2404 More actions · Answer 1

Hi Scuby,

Try this.

select REPLACE(SUBSTRING(Your_Column,1,CHARINDEX('@',Your_Column)),'@','') Name from Your_Table NOLOCK

Thanks

sibi.

Sean Lange SSC Guru Points: 286590 More actions · Answer 2

Here are a couple of other ways.

;with MyEmail (email)

as

(

select 'pearson.charles@companyname.com'

)

select *

, LEFT(email, charindex('@', email, 0) - 1)

, SUBSTRING(email, 1, CHARINDEX('@', email) - 1)

, REPLACE(SUBSTRING(email,1,CHARINDEX('@',email)),'@','')

from MyEmail

_______________________________________________________________

Need help? Help us help you.

Read the article at http://www.sqlservercentral.com/articles/Best+Practices/61537/ for best practices on asking questions.

Need to split a string? Try Jeff Modens splitter http://www.sqlservercentral.com/articles/Tally+Table/72993/.

Cross Tabs and Pivots, Part 1 – Converting Rows to Columns - http://www.sqlservercentral.com/articles/T-SQL/63681/
Cross Tabs and Pivots, Part 2 - Dynamic Cross Tabs - http://www.sqlservercentral.com/articles/Crosstab/65048/
Understanding and Using APPLY (Part 1) - http://www.sqlservercentral.com/articles/APPLY/69953/
Understanding and Using APPLY (Part 2) - http://www.sqlservercentral.com/articles/APPLY/69954/

Sean Lange SSC Guru Points: 286590 More actions · Answer 3

p.shabbir (10/25/2013)
Hi Scuby,
Try this.
select REPLACE(SUBSTRING(Your_Column,1,CHARINDEX('@',Your_Column)),'@','') Name from Your_Table NOLOCK
Thanks
sibi.

Why NOLOCK???

_______________________________________________________________

Need help? Help us help you.

Read the article at http://www.sqlservercentral.com/articles/Best+Practices/61537/ for best practices on asking questions.

Need to split a string? Try Jeff Modens splitter http://www.sqlservercentral.com/articles/Tally+Table/72993/.

Cross Tabs and Pivots, Part 1 – Converting Rows to Columns - http://www.sqlservercentral.com/articles/T-SQL/63681/
Cross Tabs and Pivots, Part 2 - Dynamic Cross Tabs - http://www.sqlservercentral.com/articles/Crosstab/65048/
Understanding and Using APPLY (Part 1) - http://www.sqlservercentral.com/articles/APPLY/69953/
Understanding and Using APPLY (Part 2) - http://www.sqlservercentral.com/articles/APPLY/69954/

p.shabbir SSCrazy Points: 2404 More actions · Answer 4

Sean Lange (10/25/2013)
p.shabbir (10/25/2013)
Hi Scuby,
Try this.
select REPLACE(SUBSTRING(Your_Column,1,CHARINDEX('@',Your_Column)),'@','') Name from Your_Table NOLOCK
Thanks
sibi.
Why NOLOCK???

Sean,

At my work i used to specify "NOLOCK" at the end of select statement.Hence, by habit i mentioned there.You can take out that.

And i checked your query.

Only this is statement is working.

REPLACE(SUBSTRING(SEmail_Addr,1,CHARINDEX('@',SEmail_Addr)),'@','').

For other two got this error.

Error : Invalid length parameter passed to the LEFT or SUBSTRING function.

Sean Lange SSC Guru Points: 286590 More actions · Answer 5

p.shabbir (10/25/2013)
Sean Lange (10/25/2013)
p.shabbir (10/25/2013)
Hi Scuby,
Try this.
select REPLACE(SUBSTRING(Your_Column,1,CHARINDEX('@',Your_Column)),'@','') Name from Your_Table NOLOCK
Thanks
sibi.
Why NOLOCK???
Sean,
At my work i used to specify "NOLOCK" at the end of select statement.Hence, by habit i mentioned there.You can take out that.

I would suggest that is a habit you learn to break unless you fully understand all the nasty stuff that hint brings to the table.

http://sqlblogcasts.com/blogs/tonyrogerson/archive/2006/11/10/1280.aspx

http://blogs.msdn.com/b/davidlean/archive/2009/04/06/sql-server-nolock-hint-other-poor-ideas.aspx

http://www.jasonstrate.com/2012/06/the-side-effect-of-nolock/[/url]

And i checked your query.
Only this is statement is working.
REPLACE(SUBSTRING(SEmail_Addr,1,CHARINDEX('@',SEmail_Addr)),'@','').
For other two got this error.
Error : Invalid length parameter passed to the LEFT or SUBSTRING function.

That's odd. The code I posted works perfectly. Now if you have email values that don't have an @ it would not work where the find code you posted does. I merely posted a couple of alternative ways. In now way did I mean that your fine code was not sufficient. In fact, I had yours in my list of possibilities.

_______________________________________________________________

Need help? Help us help you.

Read the article at http://www.sqlservercentral.com/articles/Best+Practices/61537/ for best practices on asking questions.

Need to split a string? Try Jeff Modens splitter http://www.sqlservercentral.com/articles/Tally+Table/72993/.

Cross Tabs and Pivots, Part 1 – Converting Rows to Columns - http://www.sqlservercentral.com/articles/T-SQL/63681/
Cross Tabs and Pivots, Part 2 - Dynamic Cross Tabs - http://www.sqlservercentral.com/articles/Crosstab/65048/
Understanding and Using APPLY (Part 1) - http://www.sqlservercentral.com/articles/APPLY/69953/
Understanding and Using APPLY (Part 2) - http://www.sqlservercentral.com/articles/APPLY/69954/

p.shabbir SSCrazy Points: 2404 More actions · Answer 6

Sean Lange (10/25/2013)
p.shabbir (10/25/2013)
Sean Lange (10/25/2013)
p.shabbir (10/25/2013)
Hi Scuby,
Try this.
select REPLACE(SUBSTRING(Your_Column,1,CHARINDEX('@',Your_Column)),'@','') Name from Your_Table NOLOCK
Thanks
sibi.
Why NOLOCK???
Sean,
At my work i used to specify "NOLOCK" at the end of select statement.Hence, by habit i mentioned there.You can take out that.
I would suggest that is a habit you learn to break unless you fully understand all the nasty stuff that hint brings to the table.
http://sqlblogcasts.com/blogs/tonyrogerson/archive/2006/11/10/1280.aspx
http://blogs.msdn.com/b/davidlean/archive/2009/04/06/sql-server-nolock-hint-other-poor-ideas.aspx
http://www.jasonstrate.com/2012/06/the-side-effect-of-nolock/[/url]
And i checked your query.
Only this is statement is working.
REPLACE(SUBSTRING(SEmail_Addr,1,CHARINDEX('@',SEmail_Addr)),'@','').
For other two got this error.
Error : Invalid length parameter passed to the LEFT or SUBSTRING function.
That's odd. The code I posted works perfectly. Now if you have email values that don't have an @ it would not work where the find code you posted does. I merely posted a couple of alternative ways. In now way did I mean that your fine code was not sufficient. In fact, I had yours in my list of possibilities.

Thanks for the links and your query is working perfectly.

I was running the statements individually.

Thanks.