在 MySQL 中查找重复记录

关键是重写这个查询，以便它可以用作子查询。

SELECT firstname, 
   lastname, 
   list.address 
FROM list
   INNER JOIN (SELECT address
               FROM   list
               GROUP  BY address
               HAVING COUNT(id) > 1) dup
           ON list.address = dup.address;

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SELECT date FROM logs group by date having count(*) >= 2

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为什么不只是 INNER JOIN 表本身？

SELECT a.firstname, a.lastname, a.address
FROM list a
INNER JOIN list b ON a.address = b.address
WHERE a.id <> b.id

如果地址可能存在两次以上，则需要 DISTINCT。

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我尝试了为这个问题选择的最佳答案，但这让我有些困惑。我实际上只需要在我的表中的一个字段上使用它。 this link 中的以下示例对我来说效果很好：

SELECT COUNT(*) c,title FROM `data` GROUP BY title HAVING c > 1;

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这不是更容易吗：

SELECT *
FROM tc_tariff_groups
GROUP BY group_id
HAVING COUNT(group_id) >1

?

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select `cityname` from `codcities` group by `cityname` having count(*)>=2

这是您要求的类似查询，它也 200% 工作且简单。享受！！！

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使用此查询按电子邮件地址查找重复用户...

SELECT users.name, users.uid, users.mail, from_unixtime(created)
FROM users
INNER JOIN (
  SELECT mail
  FROM users
  GROUP BY mail
  HAVING count(mail) > 1
) dupes ON users.mail = dupes.mail
ORDER BY users.mail;

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我们可以发现重复项也取决于多个字段。对于这些情况，您可以使用以下格式。

SELECT COUNT(*), column1, column2 
FROM tablename
GROUP BY column1, column2
HAVING COUNT(*)>1;

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查找重复地址比看起来要复杂得多，尤其是在您需要准确性的情况下。在这种情况下，一个 MySQL 查询是不够的......

我在 SmartyStreets 工作，我们在那里解决验证和重复数据删除等问题，我已经看到了许多具有类似问题的不同挑战。

有几个第三方服务会为您在列表中标记重复项。仅使用 MySQL 子查询执行此操作不会考虑地址格式和标准的差异。 USPS（针对美国地址）有一些制定这些标准的指导方针，但只有少数供应商获得了执行此类操作的认证。

因此，我建议您的最佳答案是将表格导出为 CSV 文件，例如，然后将其提交给有能力的列表处理器。 LiveAddress 就是其中之一，它将在几秒钟到几分钟内自动为您完成。它将使用名为“Duplicate”的新字段和其中的 Y 值来标记重复行。

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另一种解决方案是使用表别名，如下所示：

SELECT p1.id, p2.id, p1.address
FROM list AS p1, list AS p2
WHERE p1.address = p2.address
AND p1.id != p2.id

在这种情况下，您真正要做的就是获取原始列表，从中创建两个假表—— p1 和 p2——，然后在地址列上执行连接（第 3 行）。第 4 行确保同一记录不会在您的结果集中多次出现（“重复的重复项”）。

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效率不会很高，但应该可以：

SELECT *
FROM list AS outer
WHERE (SELECT COUNT(*)
        FROM list AS inner
        WHERE inner.address = outer.address) > 1;

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这将在一个表传递中选择重复项，没有子查询。

SELECT  *
FROM    (
        SELECT  ao.*, (@r := @r + 1) AS rn
        FROM    (
                SELECT  @_address := 'N'
                ) vars,
                (
                SELECT  *
                FROM
                        list a
                ORDER BY
                        address, id
                ) ao
        WHERE   CASE WHEN @_address <> address THEN @r := 0 ELSE 0 END IS NOT NULL
                AND (@_address := address ) IS NOT NULL
        ) aoo
WHERE   rn > 1

此查询实际上模拟了 Oracle 和 SQL Server 中的 ROW_NUMBER()

详情请看我博客中的文章：

分析函数：SUM、AVG、ROW_NUMBER - 在 MySQL 中模拟。

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这还将向您显示有多少重复项，并将在没有连接的情况下对结果进行排序

SELECT  `Language` , id, COUNT( id ) AS how_many
FROM  `languages` 
GROUP BY  `Language` 
HAVING how_many >=2
ORDER BY how_many DESC

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 SELECT firstname, lastname, address FROM list
 WHERE 
 Address in 
 (SELECT address FROM list
 GROUP BY address
 HAVING count(*) > 1)

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select * from table_name t1 inner join (select distinct <attribute list> from table_name as temp)t2 where t1.attribute_name = t2.attribute_name

对于您的桌子，它将类似于

select * from list l1 inner join (select distinct address from list as list2)l2 where l1.address=l2.address

此查询将为您提供列表表中所有不同的地址条目...如果您有任何主键值作为名称等，我不确定这将如何工作。

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最快的重复删除查询程序：

/* create temp table with one primary column id */
INSERT INTO temp(id) SELECT MIN(id) FROM list GROUP BY (isbn) HAVING COUNT(*)>1;
DELETE FROM list WHERE id IN (SELECT id FROM temp);
DELETE FROM temp;

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就个人而言，这个查询已经解决了我的问题：

SELECT `SUB_ID`, COUNT(SRV_KW_ID) as subscriptions FROM `SUB_SUBSCR` group by SUB_ID, SRV_KW_ID HAVING subscriptions > 1;

该脚本的作用是显示在表中多次存在的所有订户 ID 以及找到的重复项的数量。

这是表格列：

| SUB_SUBSCR_ID | int(11)     | NO   | PRI | NULL    | auto_increment |
| MSI_ALIAS     | varchar(64) | YES  | UNI | NULL    |                |
| SUB_ID        | int(11)     | NO   | MUL | NULL    |                |    
| SRV_KW_ID     | int(11)     | NO   | MUL | NULL    |                |

希望对您也有帮助！

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SELECT t.*,(select count(*) from city as tt where tt.name=t.name) as count FROM `city` as t where (select count(*) from city as tt where tt.name=t.name) > 1 order by count desc

用您的表替换城市。将名称替换为您的字段名称

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    SELECT *
    FROM (SELECT  address, COUNT(id) AS cnt
    FROM list
    GROUP BY address
    HAVING ( COUNT(id) > 1 ))

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我使用以下内容：

SELECT * FROM mytable
WHERE id IN (
  SELECT id FROM mytable
  GROUP BY column1, column2, column3
  HAVING count(*) > 1
)

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    Find duplicate Records:

    Suppose we have table : Student 
    student_id int
    student_name varchar
    Records:
    +------------+---------------------+
    | student_id | student_name        |
    +------------+---------------------+
    |        101 | usman               |
    |        101 | usman               |
    |        101 | usman               |
    |        102 | usmanyaqoob         |
    |        103 | muhammadusmanyaqoob |
    |        103 | muhammadusmanyaqoob |
    +------------+---------------------+

    Now we want to see duplicate records
    Use this query:


   select student_name,student_id ,count(*) c from student group by student_id,student_name having c>1;

+--------------------+------------+---+
| student_name        | student_id | c |
+---------------------+------------+---+
| usman               |        101 | 3 |
| muhammadusmanyaqoob |        103 | 2 |
+---------------------+------------+---+

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SELECT id, count(*) as c  
 FROM 'list'
GROUP BY id HAVING c > 1

这将返回您的 id 重复 id 的次数，或者什么都没有，在这种情况下您将没有重复的 id。

将组中的 id 更改为（例如：地址），它将返回由第一个找到的具有该地址的 id 标识的地址重复的次数。

SELECT id, count(*) as c  
 FROM 'list'
GROUP BY address HAVING c > 1

我希望它有所帮助。享受 ;）

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要快速查看重复的行，您可以运行一个简单的查询

在这里，我正在查询表并列出所有具有相同 user_id、market_place 和 sku 的重复行：

select user_id, market_place,sku, count(id)as totals from sku_analytics group by user_id, market_place,sku having count(id)>1;

要删除重复的行，您必须决定要删除哪一行。例如，具有较低 id（通常较旧）或其他一些日期信息的那个。就我而言，我只想删除较低的 id，因为较新的 id 是最新信息。

首先仔细检查是否会删除正确的记录。在这里，我在将被删除的重复项中选择记录（通过唯一 ID）。

select a.user_id, a.market_place,a.sku from sku_analytics a inner join sku_analytics b where a.id< b.id and a.user_id= b.user_id and a.market_place= b.market_place and a.sku = b.sku;

然后我运行删除查询来删除欺骗：

delete a from sku_analytics a inner join sku_analytics b where a.id< b.id and a.user_id= b.user_id and a.market_place= b.market_place and a.sku = b.sku;

备份，仔细检查，验证，验证备份然后执行。

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SELECT * FROM bookings WHERE DATE(created_at) = '2022-01-11' AND code IN (SELECT code FROM bookings GROUP BY code HAVING COUNT(code) > 1 ) 按 id DESC 排序

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当您有多个重复结果和/或当您有多个列来检查重复时，这里的大多数答案都无法解决。在这种情况下，您可以使用此查询来获取所有重复的 id：

SELECT address, email, COUNT(*) AS QUANTITY_DUPLICATES, GROUP_CONCAT(id) AS ID_DUPLICATES
    FROM list
    GROUP BY address, email
    HAVING COUNT(*)>1;

https://i.stack.imgur.com/ZQG7C.png

如果要将每个结果列为单行，则需要更复杂的查询。这是我发现的工作：

CREATE TEMPORARY TABLE IF NOT EXISTS temptable AS (    
    SELECT GROUP_CONCAT(id) AS ID_DUPLICATES
    FROM list
    GROUP BY address, email
    HAVING COUNT(*)>1
); 
SELECT d.* 
    FROM list AS d, temptable AS t 
    WHERE FIND_IN_SET(d.id, t.ID_DUPLICATES) 
    ORDER BY d.id;

https://i.stack.imgur.com/ZycST.png

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select address from list where address = any (select address from (select address, count(id) cnt from list group by address having cnt > 1 ) as t1) order by address

内部子查询返回具有重复地址的行，然后外部子查询返回具有重复地址的地址列。外部子查询必须只返回一列，因为它用作运算符 '= any' 的操作数

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Powerlord answer 确实是最好的，我建议再做一项更改：使用 LIMIT 确保 db 不会过载：

SELECT firstname, lastname, list.address FROM list
INNER JOIN (SELECT address FROM list
GROUP BY address HAVING count(id) > 1) dup ON list.address = dup.address
LIMIT 10

如果没有 WHERE 并且在进行连接时使用 LIMIT 是一个好习惯。从小值开始，检查查询的重量，然后增加限制。

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