Mysql 查找列值重复/相似的行
我想从下表中选择fname列中与第一行具有相似值的所有行。从这个表中,我想检索ID为2、5和7的行(因为“anna”在“anna”之后,“michaela”和“michaal”在“michael”之后) 到目前为止,我得到的是:Mysql 查找列值重复/相似的行,mysql,select,duplicates,Mysql,Select,Duplicates,我想从下表中选择fname列中与第一行具有相似值的所有行。从这个表中,我想检索ID为2、5和7的行(因为“anna”在“anna”之后,“michaela”和“michaal”在“michael”之后) 到目前为止,我得到的是: select *, count(fname) cnt from users group by soundex(fname) having count(soundex(fname)) > 1; 但由于我将其分组,结果是 +----+----------+---
select *, count(fname) cnt
from users group by soundex(fname)
having count(soundex(fname)) > 1;
但由于我将其分组,结果是
+----+----------+----------+-----+
| id | fname | lname | cnt |
+----+----------+----------+-----+
| 1 | anna | milski | 2 |
| 3 | michael | michaels | 3 |
+----+----------+----------+-----+
我想要的是:
+----+----------+----------+-----+
| id | fname | lname | cnt |
+----+----------+----------+-----+
| 2 | anna | nyugen | 2 |
| 5 | michaela | king | 3 |
| 7 | michaal | hardy | 3 |
+----+----------+----------+-----+
关于查询,我应该更改什么?我尝试删除“分组依据”,但它改变了结果(我可能错了,没有对其进行广泛测试)。您似乎得到了您想要的-
SOUNDEX(fname)
将只对名字而不是整个字符串进行SOUNDEX哈希。您可以研究以下几个选项:
SELECT *, COUNT(SOUNDEX(CONCAT(fname, lname))) AS cnt
GROUP BY SOUNDEX(CONCAT(fname, lname))
HAVING cnt > 1;
这取决于您想要实现什么:计算相似的名字、姓氏或两者的一些synth散列。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
);
这有点太复杂了,但它工作起来并完全满足了您的要求:)这并不能解决我的问题,请重新阅读问题。您确定您将在soundex上同时获得
michaela
和Michael
?我怀疑,你会得到他们中的任何一个。那没关系,如果它困扰你,你可以忽略它。呜呜!就是这样。你太棒了!非常感谢你。我真的很感激,这肯定会帮我在工作中省下$a:)顺便说一句,我自己也试过,但没有成功。我尝试完全删除“AND id NOT IN”子句,而是在某个地方提供带有“id>min(id)”的查询,这样它只会按顺序返回第二行(这是我想要的,看起来不那么复杂)。你们知道怎么做吗?若你们使用的是GROUPBY,你们不能只返回一行(“聚合的”)行来绕过GROUP。通过使用id>MIN(id)
您仍然只能得到一行,但这次是第二行,而不是第一行。
SELECT *, COUNT(SOUNDEX(fname)) AS cnt1, COUNT(SOUNDEX(lname)) AS cnt2
GROUP BY SOUNDEX(fname), SOUNDEX(lname)
HAVING cnt1 > 1 OR cnt2 > 1
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
);