INF: DIFFERENCE() and SOUNDEX() Functions in Transact-SQL

  - Microsoft SQL Server version 4.2 for OS/2
  - Microsoft SQL Server version 4.2

SUMMARY

The DIFFERENCE() and SOUNDEX() functions in Transact-SQL allow searches on character strings that "sound similar." SOUNDEX() converts each string into a 4-digit code. DIFFERENCE() can then be used to evaluate the level of similarity between the soundexes for two strings as returned by SOUNDEX(). For example, these functions could be used in a case where you wanted to look at all rows that sound like "Erickson," so it should find those with "Erickson," "Erikson," "Ericson," "Ericksen," "Ericsen," and so on.

MORE INFORMATION

The algorithm for determining the level of similarity between two character strings is outlined below:

given: level = difference(a1, a2)

        then
                sx1 = soundex(a1)
                sx2 = soundex(a2)
                where soundex returns:
                <alpha><numeric><numeric><numeric>

• if all characters in sx2 match all characters in sx1 (position respective) then level = 4.

• if <alpha> of sx1 is same as <alpha> of sx2, then starting level is 1; otherwise starting level is 0. Now, looping through sx1 and sx2, the level starts to "grow" by comparing one character in sx2 to all characters in sx1. If there is a match, then we increment the level and the next scan on sx1 will start from the location of the match. If no match exists, we compare the next character in sx2 to the entire 4 character list of sx1. Thus, our pointer in sx2 moves along one character at a time for every iteration and the pointer to sx1 always starts at the beginning of sx1 except in the case of a match and the location of the match becomes the starting point in sx1 for the next iteration. Whenever there is a match the level is increased. A 4 means all of the soundex characters are the same in both strings (location and value). A value of 0 means that there was no value match.

sx1     sx2

A120    A102 <-- soundex values
....    ....
0123    0123 <-- character position

iteration 2: start compare with sx2[1] and sx1[1] because of match sx2[1] == sx1[1] so level is now 2

iteration 3: start compare with sx2[2] and sx1[2] because of match sx2[2] == sx1[3] so level is now 3

We have now run out of characters in sx1 as a match was on the last character of sx1 so difference returns a value of 3.

Some more examples (remember we are always comparing sx2 to sx1):

sx1     sx2     difference

A123    A123    4
A123    B123    3
A321    A123    2
^  ^    ^^
|--+----||

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