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fralo
Posting Yak Master
161 Posts |
 Posted - 2012-01-25 : 17:52:03
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| Any other day I would know how to do this.I'm trying to figure out how to get all characters left of the '@' in an email field.select email from tableThanks for your help. |
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webfred
Master Smack Fu Yak Hacker
8781 Posts |
Posted - 2012-01-25 : 18:36:29
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select left('name@domain.com',charindex('@','name@domain.com')-1)select left(email,charindex('@',email)-1) from table
No, you're never too old to Yak'n'Roll if you're too young to die. |
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yosiasz
Master Smack Fu Yak Hacker
1635 Posts |
Posted - 2012-01-25 : 18:42:51
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| orselect substring('name@domain.com',1, charindex('@','name@domain.com')-1)If you don't have the passion to help people, you have no passion |
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visakh16
Very Important crosS Applying yaK Herder
52326 Posts |
Posted - 2012-01-26 : 09:12:10
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| i hope email values are consistent with @ included in them, if not you need a case check to see if its present or not otherwise substring will break.select substring(email,1,case when charindex('@',email)>0 then charindex('@',email) else len(email)-1 end)------------------------------------------------------------------------------------------------------SQL Server MVPhttp://visakhm.blogspot.com/ |
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fralo
Posting Yak Master
161 Posts |
Posted - 2012-01-26 : 10:05:12
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| That worked. Thanks so much for your help.I only had to add NULLIF.SELECT left(a.email,nullif(charindex('@',email)-1,-1)) |
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Query leftmost characters of email field