"Did you mean" feature on a dictionary database - php

Question

I have a ~300.000 row table; which includes technical terms; queried using PHP and MySQL + FULLTEXT indexes. But when I searching a wrong typed term; for example "hyperpext"; naturally giving no results.
I need to "compansate" little writing errors and getting nearest record from database. How I can accomplish such feaure? I know about Levenshtein distance, Soundex and Metaphone algorithms but currently not having a solid idea to implement this to querying against database.
Thanks

Answer 1

See this article for how you might implement Levenshtein distance in a MySQL stored function.
For posterity, the author's suggestion is to do this:
CREATE FUNCTION LEVENSHTEIN (s1 VARCHAR(255), s2 VARCHAR(255))
RETURNS INT
DETERMINISTIC
BEGIN
DECLARE s1_len, s2_len, i, j, c, c_temp, cost INT;
DECLARE s1_char CHAR;
DECLARE cv0, cv1 VARBINARY(256);
SET s1_len = CHAR_LENGTH(s1), s2_len = CHAR_LENGTH(s2), cv1 = 0x00, j = 1, i = 1, c = 0;
IF s1 = s2 THEN
RETURN 0;
ELSEIF s1_len = 0 THEN
RETURN s2_len;
ELSEIF s2_len = 0 THEN
RETURN s1_len;
ELSE
WHILE j <= s2_len DO
SET cv1 = CONCAT(cv1, UNHEX(HEX(j))), j = j + 1;
END WHILE;
WHILE i <= s1_len DO
SET s1_char = SUBSTRING(s1, i, 1), c = i, cv0 = UNHEX(HEX(i)), j = 1;
WHILE j <= s2_len DO
SET c = c + 1;
IF s1_char = SUBSTRING(s2, j, 1) THEN SET cost = 0; ELSE SET cost = 1; END IF;
SET c_temp = CONV(HEX(SUBSTRING(cv1, j, 1)), 16, 10) + cost;
IF c > c_temp THEN SET c = c_temp; END IF;
SET c_temp = CONV(HEX(SUBSTRING(cv1, j+1, 1)), 16, 10) + 1;
IF c > c_temp THEN SET c = c_temp; END IF;
SET cv0 = CONCAT(cv0, UNHEX(HEX(c))), j = j + 1;
END WHILE;
SET cv1 = cv0, i = i + 1;
END WHILE;
END IF;
RETURN c;
END
He also supplies a LEVENSHTEIN_RATIO helper method which will evaluate the ratio of different/total characters, rather than a straight edit distance. For instance, if it's 60%, then three-fifths of the characters in the source word are different from the destination word.
CREATE FUNCTION LEVENSHTEIN_RATIO (s1 VARCHAR(255), s2 VARCHAR(255))
RETURNS INT
DETERMINISTIC
BEGIN
DECLARE s1_len, s2_len, max_len INT;
SET s1_len = LENGTH(s1), s2_len = LENGTH(s2);
IF s1_len > s2_len THEN SET max_len = s1_len; ELSE SET max_len = s2_len; END IF;
RETURN ROUND((1 - LEVENSHTEIN(s1, s2) / max_len) * 100);
END

Answer 2

From the comments of http://dev.mysql.com/doc/refman/5.0/en/udf-compiling.html
now i download the package from the mysql udf repository
http://empyrean.lib.ndsu.nodak.edu/~nem/mysql/
wget http://empyrean.lib.ndsu.nodak.edu/~nem/mysql/udf/dludf.cgi?ckey=28
ll
tar -xzvf dludf.cgi\?ckey\=28
gcc -shared -o libmysqllevenshtein.so mysqllevenshtein.cc -I/usr/include/mysql/
mv libmysqllevenshtein.so /usr/lib
mysql -uroot -pPASS
mysql> use DATABASE
mysql> CREATE FUNCTION levenshtein RETURNS INT SONAME 'libmysqllevenshtein.so';
mysql> select levenshtein(w1.word,w2.word) as dist from word w1, word w2 where ETC........... order by dist asc limit 0,10;

Answer 3

I suggest that you generate typo variations on the query input.
i.e. hyperpext > { hyperpeext, hipertext, ... } etc
One of these is bound to be the correct spelling (especially for common misspellings)
The way you identify the most likely match is to do a lookup for each on an index which tells you the document frequency of the term. (make sense?)

Answer 4

Why not add a table column for storing the word in its alternate (e.g., Soundex) form? that way, if your first SELECT does not find the exact match, you can do a second search to look for matching alternate forms.
The trick is to encode each word so that misspelled variations end up converted into the same alternate form.