我有两张表，我想获取所有拼写相似的学校或同一所学校的学校。例如:

我的表_a:

学校

Olde School        
New School    
Other, C.S. School   
Main School
Too Cool for School

我的表_b:

学校

Old School
New ES    
Other School 
Main School  
Hardknocks School

是否可以编写一个 SELECT 查询来查找两个表中拼写相似的学校。有没有办法在列上使用 LIKE 或通配符？

例如:

SELECT  my_table_a.school, my_table_b.school
FROM ` my_table_a` ,  my_table_b
WHERE  my_table_a.directory_school_name_09_10 LIKE  my_table_b.school

我用我的真实表尝试了上面的语句，我只是得到了 '=' 会产生的结果。

基本上，我想获取每个表列中的前 4 所学校。 (当然，在现实世界中，我不会知道前4所学校是相似的)。

我正在尝试做的事情是否可行？

最佳答案

对于 Levenshtein Distance 的 UDF 实现您可能想查看“codejanitor.com: Levenshtein Distance as a MySQL Stored Function”的算法:

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;
  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

现在让我们使用您在问题中提供的数据构建一个测试用例:

CREATE TABLE table_a (name varchar(20));
CREATE TABLE table_b (name varchar(20));

INSERT INTO table_a VALUES('Olde School');      
INSERT INTO table_a VALUES('New School');
INSERT INTO table_a VALUES('Other, C.S. School');
INSERT INTO table_a VALUES('Main School');
INSERT INTO table_a VALUES('Too Cool for School');

INSERT INTO table_b VALUES('Old School');
INSERT INTO table_b VALUES('New ES');
INSERT INTO table_b VALUES('Other School');
INSERT INTO table_b VALUES('Main School');
INSERT INTO table_b VALUES('Hardknocks School');

然后:

SELECT     *
FROM       table_a a
LEFT JOIN  table_b b ON (a.name = b.name);

显然会返回学校名称完全匹配的匹配项:

+---------------------+-------------+
| name                | name        |
+---------------------+-------------+
| Olde School         | NULL        |
| New School          | NULL        |
| Other, C.S. School  | NULL        |
| Main School         | Main School |
| Too Cool for School | NULL        |
+---------------------+-------------+
5 rows in set (0.00 sec)

现在我们可以尝试使用 LEVENSHTEIN函数返回具有 edit distance 的学校名称不超过 2 个字符:

SELECT     *
FROM       table_a a
LEFT JOIN  table_b b ON (LEVENSHTEIN(a.name, b.name) <= 2);

+---------------------+-------------+
| name                | name        |
+---------------------+-------------+
| Olde School         | Old School  |
| New School          | NULL        |
| Other, C.S. School  | NULL        |
| Main School         | Main School |
| Too Cool for School | NULL        |
+---------------------+-------------+
5 rows in set (0.08 sec)

现在使用 <= 3作为编辑距离阈值:

SELECT     *
FROM       table_a a
LEFT JOIN  table_b b ON (LEVENSHTEIN(a.name, b.name) <= 3);

我们得到以下结果:

+---------------------+--------------+
| name                | name         |
+---------------------+--------------+
| Olde School         | Old School   |
| Olde School         | Other School |
| New School          | Old School   |
| Other, C.S. School  | NULL         |
| Main School         | Main School  |
| Too Cool for School | NULL         |
+---------------------+--------------+
6 rows in set (0.06 sec)

请注意这次如何 Olde School也匹配 Other School , 和 New School匹配Old School以及。这些可能是误报，表明定义阈值对于避免错误匹配非常重要。

解决此问题的一种常用技术是在应用阈值时考虑字符串的长度。事实上，site that I cited for this implementation还提供了一个LEVENSHTEIN_RATIO返回基于字符串长度的编辑差异的比率(百分比)的函数。

关于mysql - 如何在 MySQL 中的两个不同表中选择相似的行(可能吗？)，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/3376082/