MySQL select maximum length of matching string

Question

I need to return all the text results(s), if any, that share the maximum length left bounded substring common to the search string.

Given a search for \"StackOverflo

Answer 1

You could do a query after creating a Levenshtein Distance stored function. This could get the best matched results for you.

This is not my code. I got this from here. It does seem to test well on sqlfiddle.

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;

Your query could then look something like this:

SELECT names, levenshtein(`names`, 'StackOverflow') as dist
FROM mytable
ORDER BY dist;

Here is what this looks like over on sqlfiddle.

The results would look like this with the lowest distance being the closest match:

NAMES           DIST
StackOverthrow  3
StackFlow       4
Soverflow       4
StackOv         6
Stack           8
Sta             10

Answer 2

With an index on name, the following ought to be extremely performant:

SELECT DISTINCT name
FROM   myTable
WHERE  name LIKE CASE
  WHEN NOT EXISTS(SELECT * FROM myTable WHERE name LIKE 'S%') THEN '%'
  WHEN NOT EXISTS(SELECT * FROM myTable WHERE name LIKE 'St%') THEN 'S%'
  WHEN NOT EXISTS(SELECT * FROM myTable WHERE name LIKE 'Sta%') THEN 'St%'
  WHEN NOT EXISTS(SELECT * FROM myTable WHERE name LIKE 'Stac%') THEN 'Sta%'
  WHEN NOT EXISTS(SELECT * FROM myTable WHERE name LIKE 'Stack%') THEN 'Stac%'
  WHEN NOT EXISTS(SELECT * FROM myTable WHERE name LIKE 'StackO%') THEN 'Stack%'
  WHEN NOT EXISTS(SELECT * FROM myTable WHERE name LIKE 'StackOv%') THEN 'StackO%'
  WHEN NOT EXISTS(SELECT * FROM myTable WHERE name LIKE 'StackOve%') THEN 'StackOv%'
  WHEN NOT EXISTS(SELECT * FROM myTable WHERE name LIKE 'StackOver%') THEN 'StackOve%'
  WHEN NOT EXISTS(SELECT * FROM myTable WHERE name LIKE 'StackOverf%') THEN 'StackOver%'
  WHEN NOT EXISTS(SELECT * FROM myTable WHERE name LIKE 'StackOverfl%') THEN 'StackOverf%'
  WHEN NOT EXISTS(SELECT * FROM myTable WHERE name LIKE 'StackOverflo%') THEN 'StackOverfl%'
  WHEN NOT EXISTS(SELECT * FROM myTable WHERE name LIKE 'StackOverflow%') THEN 'StackOverflo%'
  ELSE 'StackOverflow%'
END

See it on sqlfiddle.

Answer 3

Don't know why you would look at the smallest first. I would do that in reverse... try by the LONGEST EXACT match first, and if not found, work backwards 1 character at a time until one is found.