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SQL extract number of a string

Posted on 2013-01-17
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Last Modified: 2013-02-22
Hello

I use this function to exctract number out of a string, it worked great until somebody put the
word 'Inc.'  instead of a number the '.' created a problem and generate an error:

declare @param varchar(50)
select Left(SubString(@param, PatIndex('%[0-9.-]%', @param), 8000), PatIndex('%[^0-9.-]%', SubString(@param, PatIndex('%[0-9.-]%', @param), 8000) + 'X')-1)

Thanks for the help
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Question by:arnololo123
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Expert Comment

by:LordKnightshade
ID: 38789843
If you only need to worry about accounting for the period just wrap the SELECT value in a REPLACE targeting the period.  It's not particularly elegant, but it works :)

declare @param varchar(50)
select @param = 'Inc.'
select REPLACE(Left(SubString(@param, PatIndex('%[0-9.-]%', @param), 8000),PatIndex('%[^0-9.-]%',
	SubString(@param, PatIndex('%[0-9.-]%', @param), 8000) + 'X')-1),'.','')

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Author Comment

by:arnololo123
ID: 38790013
Well the problem with this approach is that the value 3.25 for example would be stripped of the period.
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Expert Comment

by:LordKnightshade
ID: 38793076
Is it feasible for you to use a second variable in the function?  If so you can cut down on the code by evaluating a variable after the evaluation.

declare @param varchar(50), @param2 varchar(50)
select @param = '3.25'
select @param2 = Left(SubString(@param, PatIndex('%[0-9.-]%', @param), 8000),PatIndex('%[^0-9.-]%',
	SubString(@param, PatIndex('%[0-9.-]%', @param), 8000) + 'X')-1)
select (CASE @param2 WHEN '.' THEN '' ELSE @param2 END)

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Accepted Solution

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Scott Pletcher earned 500 total points
ID: 38793825
declare @param varchar(50)
select Left(SubString(@param, PatIndex('%[^a-z][0-9.-]%', @param), 8000), PatIndex('%[^0-9.-]%', SubString(@param, PatIndex('%[^a-z][0-9.-]%', @param), 8000) + 'X')-1)
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Expert Comment

by:mimran18
ID: 38822112
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