Question

我想查找姓名与声音匹配的联系人表格中的所有重复名称。例如：Rita或Reeta，Microsoft或Microsift，Mukherjee或Mukherji。

我使用了以下查询：

SELECT contacts.id 
FROM contacts 
INNER JOIN (
    SELECT first_name, last_name, count(*) AS rows 
    FROM contacts 
    WHERE deleted = 0 
    GROUP BY SOUNDEX(first_name), SOUNDEX(last_name) 
    HAVING count(rows) > 1
) AS p 
WHERE contacts.deleted = 0 
AND p.first_name SOUNDS LIKE contacts.first_name 
AND p.last_name SOUNDS LIKE contacts.last_name 
ORDER BY contacts.date_entered DESC

上述查询给出了正确的结果，但是当记录很多时会花费很多时间。

Answer 1

我不知道比SOUNDEX()更好的（原生）方法。它之所以慢，是因为它是一个功能，因此需要处理所有记录以计算价值并从那里开始工作。解决这个问题的方法是将结果直接存储到表中。我没有在MySQL中使用这些函数的经验，但根据documentation，您似乎可以将WHERE子句转换为

[...] AND SOUNDEX(p.first_name) = SOUNDEX(contacts.first_name) [...]

因此，如果您已预先计算（并编制索引！）这些值，则搜索匹配记录的速度应该更快！

那说我很难搞清楚你的查询。我认为你不需要那里的HAVING COUNT(*) > 1，即便如此，我对你想如何分组/过滤联系人感到困惑！？

你想要这样的东西：

SELECT c1.id as contact_id, 
       c2.id as similar_id
  FROM contacts c1 
  JOIN contacts c2
    ON c2.id <> c1.id
   AND c2.deleted = 0
   AND SOUNDEX(c2.first_name) = SOUNDEX(c1.first_name)
   AND SOUNDEX(c2.last_name) = SOUNDEX(c1.last_name)
 WHERE c1.deleted = 0 
ORDER BY c1.date_entered DESC

然后您可以根据上面的建议选择

SELECT c1.id as contact_id, 
       c2.id as similar_id
  FROM contacts c1 
  JOIN contacts c2
    ON c2.id <> c1.id
   AND c2.deleted = 0
   AND c2.first_name_soundex = c1.first_name_soundex
   AND c2.last_name_soundex = c1.last_name_soundex
 WHERE c1.deleted = 0 
ORDER BY c1.date_entered DESC

其中first_name_soundex包含SOUNDEX（first_name）等的结果。建立索引时，您可能希望覆盖索引超过deleted，first_name_soundex，last_name_soundex。（AFAIK MySQL尚未支持过滤索引，否则您可以仅将索引限制为deleted = 0。

Answer 2

SOUNDEX是（恕我直言）实用性非常有限。一个极端的例子......

SELECT SOUNDEX('cholmondley');
+------------------------+
| SOUNDEX('cholmondley') |
+------------------------+
| C4534                  |
+------------------------+

SELECT SOUNDEX('chumleigh');
+----------------------+
| SOUNDEX('chumleigh') |
+----------------------+
| C542                 |
+----------------------+

获取类似声音的记录

2 个答案: