查找具有重复/相似列值的行 MySQL

Question

我想从下表中选择 fname 列中具有相似值的所有行作为其顺序中的第一行。 IOW 从此表中我想检索 id 为 2,5 和 7 的行（因为“ anna”位于“anna”和“michaela”之后>”和“michaal”位于“michael”之后）。

+----+------------+----------+
| id | fname      | lname    |
+----+------------+----------+
|  1 | anna       | milski   |
|  2 |  anna      | nguyen   |
|  3 | michael    | michaels |
|  4 | james      | bond     |
|  5 | michaela   | king     |
|  6 | bruce      | smart    |
|  7 | michaal    | hardy    |
+----+------------+----------+

到目前为止我所拥有的是这样的：

select *, count(fname) cnt 
from users group by soundex(fname) 
having count(soundex(fname)) > 1;

但由于我将其分组，所以结果是

+----+----------+----------+-----+
| id | fname    | lname    | cnt |
+----+----------+----------+-----+
|  1 | anna     | milski   |   2 |
|  3 | michael  | michaels |   3 |
+----+----------+----------+-----+

我想要检索的是这样的：

+----+----------+----------+-----+
| id | fname    | lname    | cnt |
+----+----------+----------+-----+
|  2 |  anna    | nyugen   |   2 |
|  5 | michaela | king     |   3 |
|  7 | michaal  | hardy    |   3 |
+----+----------+----------+-----+

我应该对查询进行哪些更改？我尝试删除“group by”，但它改变了结果（我可能是错的，没有广泛测试它）。

Answer 1

我重新阅读了您最初的问题，并提出了以下解决方案：

SELECT *
FROM   users
WHERE  id IN
       (SELECT id
       FROM    users t4
               INNER JOIN
                       (SELECT  soundex(fname) AS snd,
                                COUNT(*)       AS cnt
                       FROM     users          AS t5
                       GROUP BY snd
                       HAVING   cnt > 1
                       )
                       AS t6
               ON      soundex(t4.fname)=snd
       )
AND    id NOT IN
       (SELECT  MIN(t2.id) AS wanted
       FROM     users t2
                INNER JOIN
                         (SELECT  soundex(fname) AS snd,
                                  COUNT(*)       AS cnt
                         FROM     users          AS t1
                         GROUP BY snd
                         HAVING   cnt > 1
                         )
                         AS t3
                ON       soundex(t2.fname)=snd
       GROUP BY snd
       );

它有点过于复杂，但它可以正常工作并完全满足您的要求:)

Answer 2

您似乎得到了您所要求的 - SOUNDEX(fname) 将使 Soundex 哈希值仅来自名字，而不是整个字符串。您可以研究一些选项：

SELECT *, COUNT(SOUNDEX(CONCAT(fname, lname))) AS cnt
GROUP BY SOUNDEX(CONCAT(fname, lname))
HAVING cnt > 1;

或者

SELECT *, COUNT(SOUNDEX(fname)) AS cnt1, COUNT(SOUNDEX(lname)) AS cnt2
GROUP BY SOUNDEX(fname), SOUNDEX(lname)
HAVING cnt1 > 1 OR cnt2 > 1

这取决于您想要实现的目标：相似的名字、姓氏或两者的一些合成哈希的计数。