

Forum Newbie
Group: General Forum Members
Last Login: Wednesday, December 16, 2015 6:46 AM
Points: 5,
Visits: 6


What exactly and on what basis does the soundex function returns the integer values.
I read through msdn but it only tells that it doesn't considers vowels in the strings.
But what I exactly want to know is what logic is used during comparison of two strings behind th scene.




SSChampion
Group: General Forum Members
Last Login: Saturday, November 14, 2015 5:53 AM
Points: 13,873,
Visits: 9,610


There's an entry in Wikipedia about soundex. It has what you're looking for.
 Gus "GSquared", RSVP, OODA, MAP, NMVP, FAQ, SAT, SQL, DNA, RNA, UOI, IOU, AM, PM, AD, BC, BCE, USA, UN, CF, ROFL, LOL, ETC Property of The Thread
"Nobody knows the age of the human race, but everyone agrees it's old enough to know better."  Anon




Forum Newbie
Group: General Forum Members
Last Login: Wednesday, December 16, 2015 6:46 AM
Points: 5,
Visits: 6


Thanks. This is exactly what I am looking for




SSC Veteran
Group: General Forum Members
Last Login: Yesterday @ 12:22 PM
Points: 272,
Visits: 1,498


Here is my first attempt to make soundex more selective:
Algorithm sqlserver uses: Replace consonants with digits as follows (but do not change the first letter): b, f, p, v => 1 c, g, j, k, q, s, x, z => 2 d, t => 3 l => 4 m, n => 5 r => 6 Collapse adjacent identical digits into a single digit of that value. Remove all nondigits after the first letter. Return the starting letter and the first three remaining digits. If needed, append zeroes to make it a letter and three digits.
Lets extend this to 5 in length to make it more selective. On lastname data I see a selectivity increase from 1 in 3000 to 1 in 10000.
Function contains no error checking and is not optimized.
drop function dbo.soundex5; go
create function Soundex5( @name varchar(30)) returns varchar(5) as begin
declare @out varchar(20), @i int, @vc varchar(1);
set @name = @name + '00000';
set @out = left(@name,1);
set @i = 2; while @i <= len(@name) begin set @vc = substring(@name, @i, 1)
set @out = @out + case when charindex(@vc, 'bfpv') > 0 then '1' when charindex(@vc, 'cgjkqsxz') > 0 then '2' when charindex(@vc, 'dt') > 0 then '3' when charindex(@vc, 'l') > 0 then '4' when charindex(@vc, 'mn') > 0 then '5' when charindex(@vc, 'r') > 0 then '6' when charindex(@vc, '0') > 0 then '0' else '' end;
set @i = @i + 1; end
 ignore similar consecutive sounds set @out = REPLACE(@out,'11','1'); set @out = REPLACE(@out,'22','2'); set @out = REPLACE(@out,'33','3'); set @out = REPLACE(@out,'44','4'); set @out = REPLACE(@out,'55','5'); set @out = REPLACE(@out,'66','6');
return left(@out,5);
end go



