3

我正在尝试创建一个用于计算 levenshtein 距离的 MySQL 函数。我找到了一个看起来非常接近我需要的函数，但是它总是到处抛出错误——我是 MySQL 函数的新手，所以我不知道出了什么问题？

这是功能：

DELIMITER $$

CREATE FUNCTION LEVENSHTEIN( s1 CHAR(255), s2 CHAR(255)) 
RETURNS int(3) 
DETERMINISTIC
BEGIN
    DECLARE s1_len, s2_len, i, j, c, c_temp, cost INT;
    DECLARE s1_char CHAR(255);
    DECLARE cv0, cv1 CHAR(255);

    SET s1_len = LENGTH(s1);
    SET s2_len = LENGTH(s2);
    SET cv1 = 0x00;
    SET j = 1;
    SET i = 1;
    SET c = 0;

    IF s1 = s2 THEN
        RETURN 0;
    ELSE IF s1_len = 0 THEN
        RETURN s2_len;
    ELSE IF s2_len = 0 THEN
        RETURN s1_len;
    ELSE
        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; 
                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$$

DELIMITER ;

这是错误（到目前为止）。

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHILE; END IF; RETURN c; END' at line 36

编辑：原始代码可以在这里找到

4

3 回答 3

4

delimiter在创建函数之前不要忘记更改。

DELIMITER $$

CREATE FUNCTION LEVENSHTEIN( s1 CHAR(255), s2 CHAR(255)) 
RETURNS int(3) 
DETERMINISTIC
BEGIN
    DECLARE s1_len, s2_len, i, j, c, c_temp, cost INT;
    DECLARE s1_char CHAR(255);
    DECLARE cv0, cv1 CHAR(255);

    SET s1_len = LENGTH(s1);
    SET s2_len = LENGTH(s2);
    SET cv1 = 0x00;
    SET j = 1;
    SET i = 1;
    SET c = 0;

    IF s1 = s2 THEN
        RETURN 0;
    ELSE IF s1_len = 0 THEN
        RETURN s2_len;
    ELSE IF s2_len = 0 THEN
        RETURN s1_len;
    ELSE
        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; 
                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$$

DELIMITER ;
于 2012-09-27T08:48:19.240 回答
4

这对我来说很好

DELIMITER $$
CREATE FUNCTION levenshtein( s1 VARCHAR(255), s2 VARCHAR(255) )
RETURNS INT
DETERMINISTIC
BEGIN
DECLARE s1_len, s2_len, i, j, c, c_temp, cost INT;
DECLARE s1_char CHAR;
-- max strlen=255
DECLARE cv0, cv1 VARBINARY(256);
SET s1_len = CHAR_LENGTH(s1), s2_len = CHAR_LENGTH(s2), cv1 = 0x00, j = 1, i = 1, c = 0;
IF s1 = s2 THEN
RETURN 0;
ELSEIF s1_len = 0 THEN
RETURN s2_len;
ELSEIF s2_len = 0 THEN
RETURN s1_len;
ELSE
WHILE j <= s2_len DO
SET cv1 = CONCAT(cv1, UNHEX(HEX(j))), j = j + 1;
END WHILE;
WHILE i <= s1_len DO
SET s1_char = SUBSTRING(s1, i, 1), c = i, cv0 = UNHEX(HEX(i)), j = 1;
WHILE j <= s2_len DO
SET c = c + 1;
IF s1_char = SUBSTRING(s2, j, 1) THEN
SET cost = 0; ELSE SET cost = 1;
END IF;
SET c_temp = CONV(HEX(SUBSTRING(cv1, j, 1)), 16, 10) + cost;
IF c > c_temp THEN SET c = c_temp; END IF;
SET c_temp = CONV(HEX(SUBSTRING(cv1, j+1, 1)), 16, 10) + 1;
IF c > c_temp THEN
SET c = c_temp;
END IF;
SET cv0 = CONCAT(cv0, UNHEX(HEX(c))), j = j + 1;
END WHILE;
SET cv1 = cv0, i = i + 1;
END WHILE;
END IF;
RETURN c;
END$$
DELIMITER ;
于 2013-12-10T12:56:25.467 回答
2

您忘记了语句分隔符 ';' 几乎在每一行的末尾。

于 2012-09-27T08:45:21.020 回答