How can I efficiently find similar terms in a MySQL database using the Levenshtein distance?

Finding Similar Terms in MySQL Using Levenshtein Distance

The Levenshtein distance is a measure of the similarity between two strings. It can be used to find similar terms in a database, which can be useful for tasks such as autocompletion and spell checking.

One way to find similar terms in MySQL is to use the levenshtein() function. This function takes two strings as input and returns the Levenshtein distance between them. The following PHP code demonstrates how to use the levenshtein() function to find similar terms in a database:

$word = strtolower($_GET['term']);

$lev = 0;

$q = mysql_query("SELECT `term` FROM `words`");
while($r = mysql_fetch_assoc($q)) 
{ 
    $r['term'] = strtolower($r['term']); 

    $lev = levenshtein($word, $r['term']);

    if($lev >= 0 && $lev <p>However, this approach can be inefficient if there are a large number of terms in the database, as it requires a separate query for each term. To improve efficiency, it is possible to use a single query to find all terms that are within a certain Levenshtein distance of the input term.</p><p>To do this, you need to use a MySQL function to calculate the Levenshtein distance. The following MySQL function can be used:</p><pre class="brush:php;toolbar:false">CREATE FUNCTION levenshtein(s1 VARCHAR(255), s2 VARCHAR(255)) RETURNS INT
BEGIN
  DECLARE s1_len INT, s2_len INT, i INT, j INT, c INT, d INT;
  SET s1_len = LENGTH(s1), s2_len = LENGTH(s2), i = 0, j = 0, c = 0, d = 0;
  IF s1_len = 0 THEN RETURN s2_len;
  ELSEIF s2_len = 0 THEN RETURN s1_len;
  END IF;
 
  DECLARE cost_matrix INT[][] DEFAULT (SELECT * FROM (
    SELECT a.i_col, b.j_row, IF(a.i_col = 0, b.j_row, IF(b.j_row = 0, a.i_col, IF(SUBSTR(s1, a.i_col, 1) = SUBSTR(s2, b.j_row, 1), 0, 1))) AS cost
    FROM (
      SELECT 1 AS i_col
      UNION ALL
      SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9 UNION ALL SELECT 10 UNION ALL SELECT 11 UNION ALL SELECT 12 UNION ALL SELECT 13 UNION ALL SELECT 14 UNION ALL SELECT 15
    ) AS a
    CROSS JOIN
    (
      SELECT 1 AS j_row
      UNION ALL
      SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9 UNION ALL SELECT 10 UNION ALL SELECT 11 UNION ALL SELECT 12 UNION ALL SELECT 13 UNION ALL SELECT 14 UNION ALL SELECT 15
    ) AS b
  ) AS subquery);
 
  WHILE i  0 THEN
        SET d = LEAST(d, cost_matrix[i][j-1] + 1);
      END IF;
      IF i > 0 THEN
        SET d = LEAST(d, cost_matrix[i-1][j-1] + c);
      END IF;
 
      SET cost_matrix[i][j] = d;
      SET j = j + 1;
    END WHILE;
    SET j = 0;
    SET i = i + 1;
  END WHILE;
 
  RETURN cost_matrix[s1_len][s2_len];
END;

Once you have created this function, you can use it to find similar terms in a database using a single query. The following query finds all terms in the words table that are within a Levenshtein distance of 4 from the input term:

$word = mysql_real_escape_string($word);
mysql_qery("SELECT `term` FROM `words` WHERE levenshtein('$word', `term`) BETWEEN 0 AND 4");

This query will return a list of all terms that are within a Levenshtein distance of 4 from the input term, sorted in ascending order of Levenshtein distance.

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