这个MySQL函数有什么问题

时间:2011-06-30 15:22:08

标签: mysql mysql-error-1064

我在这个链接中看到了这个功能:

http://www.artfulsoftware.com/infotree/queries.php#552

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;

但是当我尝试执行时,我收到此错误消息:

1064 - 您的SQL语法出错;检查与MySQL服务器版本对应的手册,以便在第5行的''附近使用正确的语法

MySQL版本:5.1.53

1 个答案:

答案 0 :(得分:2)

符!

DELIMITER $$

CREATE FUNCTION levenshtein( s1 VARCHAR(255), s2 VARCHAR(255) )
  RETURNS INT
  DETERMINISTIC
  BEGIN
  ...
  ...
  ...
  END$$

DELIMITER ;