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Question

Ich bin kein Experte SQL, aber ich möchte die Abfrage wissen, die zur gleichen Zeit sucht gibt diese Regeln

# 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. 

Etwas wie:

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

Sie benötigen eine Implementierung von Levenshtein. Wie diese: 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; // 

Dann können Sie es wie folgt verwenden:

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

Answer 2

DELIMITER IST ERFORDERLICH FÜR SIE ARBEITEN

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 //