慕姐8265434
上面给出的levenshtein <= 1的函数不正确 - 它给出了不正确的结果，例如“床”和“出价”。我在第一个答案中修改了上面给出的“MySQL Levenshtein距离查询”，以接受一个“限制”，这将加快它的速度。基本上，如果你只关心Levenshtein <= 1，将极限设置为“2”，如果它是0或1，函数将返回精确的levenshtein距离; 如果精确的levenshtein距离为2或更大，则为2。这个mod使它快15％到50％ - 搜索词越长，优势越大（因为算法可以提前保释。）例如，搜索200,000个单词以查找距离1的所有匹配项“傻笑”，原版在我的笔记本电脑上花了3分47秒，而“极限”版本需要1:39。当然，这些对于任何实时使用来说都太慢了。码：DELIMITER $$CREATE FUNCTION levenshtein_limit_n( s1 VARCHAR(255), s2 VARCHAR(255), n INT)
RETURNS INT
DETERMINISTIC
BEGIN
DECLARE s1_len, s2_len, i, j, c, c_temp, cost, c_min INT;
DECLARE s1_char CHAR;
-- max strlen=255
DECLARE cv0, cv1 VARBINARY(256);
SET s1_len = CHAR_LENGTH(s1), s2_len = CHAR_LENGTH(s2), cv1 = 0x00, j = 1, i = 1, c = 0, c_min = 0;
IF s1 = s2 THEN
RETURN 0;
ELSEIF s1_len = 0 THEN
RETURN s2_len;
ELSEIF s2_len = 0 THEN
RETURN s1_len;
ELSE
WHILE j <= s2_len DO
SET cv1 = CONCAT(cv1, UNHEX(HEX(j))), j = j + 1;
END WHILE;
WHILE i <= s1_len and c_min < n DO -- if actual levenshtein dist >= limit, don't bother computing it
SET s1_char = SUBSTRING(s1, i, 1), c = i, c_min = i, cv0 = UNHEX(HEX(i)), j = 1;
WHILE j <= s2_len DO
SET c = c + 1;
IF s1_char = SUBSTRING(s2, j, 1) THEN
SET cost = 0; ELSE SET cost = 1;
END IF;
SET c_temp = CONV(HEX(SUBSTRING(cv1, j, 1)), 16, 10) + cost;
IF c > c_temp THEN SET c = c_temp; END IF;
SET c_temp = CONV(HEX(SUBSTRING(cv1, j+1, 1)), 16, 10) + 1;
IF c > c_temp THEN
SET c = c_temp;
END IF;
SET cv0 = CONCAT(cv0, UNHEX(HEX(c))), j = j + 1;
IF c < c_min THEN
SET c_min = c;
END IF;
END WHILE;
SET cv1 = cv0, i = i + 1;
END WHILE;
END IF;
IF i <= s1_len THEN -- we didn't finish, limit exceeded
SET c = c_min; -- actual distance is >= c_min (i.e., the smallest value in the last computed row of the matrix)
END IF;
RETURN c;
END$$