mysql - 智能搜索MySQL

Question

我不是 SQL 专家，但我想知道同时搜索的查询是否通过了这些规则

# they differ only by one symbol;
# user's input has one extra symbol;
# user's input has one missing symbol.
# Note that the comparison should be case-insensitive.

类似于:

SELECT `name` FROM `table` WHERE `name` = 'frd'

| name |
--------
  ford <---- # user's input has one missing symbol.
  crd <----- # they differ only by one symbol
  fr <------ # user's input has one extra symbol;
  FRD <----- # Note that the comparison should be case-insensitive.

Answer 1

您需要 Levenshtein 的实现。像这个:http://www.artfulsoftware.com/infotree/qrytip.php?id=552

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; //

然后你可以像这样使用它:

SELECT * FROM yourTable WHERE levenshtein(yourColumn, 'word') <= 1