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Select just three octets from IP number


Select just three octets from IP number

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rodrigo.rollo
rodrigo.rollo
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I need to create a function to select just 3 octets from a column (ip_number).
I paste below the select that I did, but it is very large. anyone have any idea so that I can reduce this select?



-- I want to identify only three octets, the IP number

declare @x varchar(20)
select @x = '10.0.42.162'
select @x = '10.0.200.120'

select
@x Ip,
substring(@x,1,len(substring(@x,1,charindex('.', @x ))))+
substring(substring(@x,len(substring(@x,1,len(substring(@x,1,charindex('.', @x )))))+1,len(@x)),1,
len(substring(substring(@x,len(substring(@x,1,len(substring(@x,1,charindex('.', @x )))))+1,len(@x)),1,
charindex('.', substring(@x,len(substring(@x,1,len(substring(@x,1,charindex('.', @x )))))+1,len(@x)) )))) +
substring(SUBSTRING(@x,len(substring(@x,1,len(substring(@x,1,charindex('.', @x ))))+
substring(substring(@x,len(substring(@x,1,len(substring(@x,1,charindex('.', @x )))))+1,len(@x)),1,
len(substring(substring(@x,len(substring(@x,1,len(substring(@x,1,charindex('.', @x )))))+1,len(@x)),1,
charindex('.', substring(@x,len(substring(@x,1,len(substring(@x,1,charindex('.', @x )))))+1,len(@x)) )))))+1,
LEN(@x)),1,charindex('.', SUBSTRING(@x,len(substring(@x,1,len(substring(@x,1,charindex('.', @x ))))+
substring(substring(@x,len(substring(@x,1,len(substring(@x,1,charindex('.', @x )))))+1,len(@x)),1,
len(substring(substring(@x,len(substring(@x,1,len(substring(@x,1,charindex('.', @x )))))+1,len(@x)),1,
charindex('.', substring(@x,len(substring(@x,1,len(substring(@x,1,charindex('.', @x )))))+1,len(@x)) )))))+1,
LEN(@x)) )) rede
Jason Selburg
Jason Selburg
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DECLARE @t VARCHAR(15)

SET @t = '111.222.333.444'

SELECT PARSENAME(@t,4),PARSENAME(@t,3),PARSENAME(@t,2)

______________________________________________________________________

Personal Motto: Why push the envelope when you can just open it?

If you follow the direction given HERE you'll likely increase the number and quality of responses you get to your question.

Jason L. Selburg
Hari.Sharma
Hari.Sharma
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DECLARE @t VARCHAR(15)
SET @t = '10.0.123.77'
SELECT PARSENAME(@t,4)+'.'+PARSENAME(@t,3)+'.'+PARSENAME(@t,2)

Cheers,
Hari
Tips & Tricks for SQL BI Developers

Hari.Sharma
Hari.Sharma
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Alternatively u can do:

DECLARE @t VARCHAR(15)
SET @t = '10.0.123.77'
SELECT Left(@t,Len(@t)-CHARINDEX('.',Reverse(@t)))

Cheers,
Hari
Tips & Tricks for SQL BI Developers

robbaker
robbaker
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This is great information, thanks!
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