mysql - 在mysql中查找相似的句子

Question

根据 Finding how similar two strings are有几种方法可以计算两个字符串之间的相似性度量。

Soundex 是一种非常糟糕的算法，目前可用于 mysql。是否有任何其他方法实现可用于 mysql？

Answer 1

当然，有一个很好的 Levenshtein distance 的 MySQL 实现:

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