How to Solve This Using Sql ?

Question

1.00/5 (1 vote)

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SQL-Server-2008R2

Stdid	Firstname		Email
1	Vivek	         	vivek@abc.com
2	Pankaj	        	pankaj@gmail.com
3	Amit	         	amit@yahoo.com
4	Manish		        manish@hotmail.com
5	Abhishek		    abhishek@abcd.com

I need my result like 

Stdid	Firstname       part1       part2   part3
1	Vivek	          vivek         @abc     .com
2	Pankaj            pankaj       @gmail    .com
3	Amit	          amit         @yahoo    .com
4	Manish		      manish       @hotmail  .com
5	Abhishek          abhishek     @abcd     .com

The part1 can be off many charcters so it must analyse the words before @ and second part must be off @followed by domain name till before. alone and '.com' must always be recoganised in third part

Posted 18-Aug-15 1:21am

Member 11917248

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4 solutions

Solution 4

Hello ,
You may use CHARINDEX and SUBSTRING to get required op.

declare @name varchar(30)
set @name='abhishek@abcd.com'


declare @a varchar(20)
declare @b varchar(20)
declare @c varchar(20)

--get the first part
select @a=substring(@name,1,charindex('@',@name)-1)

--get the second part
select @b=substring(@name,charindex('@',@name),charindex('.',@name)-charindex('@',@name))
--get the third part

select @c=substring(@name,charindex('.',@name),len(@name)-charindex('.',@name)+1)

select @a,@b,@c

Posted 18-Aug-15 1:48am

Animesh Datta

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OriginalGriff · Accepted Answer · 2015-08-18T01:44:00

Try:

SQL

SELECT  stdid, 
        Firstname, 
        SUBSTRING(Email, 1, CHARINDEX('@', Email,1) - 1) AS Part1,
        SUBSTRING(Email, CHARINDEX('@', Email,1), CHARINDEX('.', Email,1) - CHARINDEX('@', Email,1)) AS Part2,
        SUBSTRING(Email, CHARINDEX('.', Email,1), 9999) AS Part3
FROM MyTable

rkganapathy · Accepted Answer · 2015-08-18T01:37:00

Solution 2

http://stackoverflow.com/questions/21883080/how-to-split-an-email-address-into-its-parts[^]

Posted 18-Aug-15 1:37am

rkganapathy

Maciej Los · Accepted Answer · 2015-08-18T11:40:00

Solution 5

Try this:

SQL

SELECT  stdid,
        Firstname,
        LEFT(Email, CHARINDEX('@', Email,1) - 1) AS Part1,
        SUBSTRING(Email, CHARINDEX('@', Email), CHARINDEX('.', Email) - CHARINDEX('@', Email)) AS Part2,
        RIGHT(Email, CHARINDEX('.', REVERSE(Email))) AS Part3
FROM MyTable

Posted 18-Aug-15 11:40am

Maciej Los