在MySQL中查找重复记录

我想在MySQL数据库中提取重复记录。这可以通过以下方式完成：

SELECT address, count(id) as cnt FROM list
GROUP BY address HAVING cnt > 1

结果是：

100 MAIN ST    2

我想将其拉出，以使其显示重复的每一行。就像是：

JIM    JONES    100 MAIN ST
JOHN   SMITH    100 MAIN ST

关于如何做到这一点的任何想法？我试图避免做第一个，然后在代码中用第二个查询查找重复项。

关键是重写此查询，以便可以将其用作子查询。

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;

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

为什么不将INNER JOIN本身与桌子相连？

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

如果地址可以存在两次以上，则需要DISTINCT。

我尝试了为该问题选择的最佳答案，但是这使我有些困惑。实际上，我只需要在表中的单个字段上使用它。以下来自此链接的示例对我来说非常有效：

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

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

这是您所要求的类似查询，并且200％的工作也很容易。请享用！！！

这难道不是很容易吗：

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

？

通过此查询通过电子邮件地址查找重复的用户...

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;

我们发现重复项还取决于多个字段。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;

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

我在SmartyStreets工作，我们致力于解决验证和重复数据删除等问题，并且我遇到了许多类似问题的各种挑战。

有一些第三方服务会为您标记列表中的重复项。仅使用MySQL子查询执行此操作不会解决地址格式和标准的差异。USPS（用于美国地址）具有制定这些标准的某些准则，但是只有少数供应商被认证可以执行此类操作。

因此，我建议您最好的答案是将表导出到CSV文件中，然后将其提交给有能力的列表处理器。SmartyStreets 批量地址验证工具就是其中之一，它将在几秒钟到几分钟内自动为您完成。它将使用称为“ Duplicate”的新字段及其中的值来标记重复的行Y。

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​​要做的只是获取原始列表表，创建两个p retend表-p 1和p 2，然后在address列上执行（第3行）。第4行确保同一条记录不会在您的结果集中多次显示（“重复重复”）。

效率不是很高，但是应该可以工作：

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

这将在一个表遍中选择重复项，而不选择子查询。

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

该查询实际上模拟ROW_NUMBER()了Oracle和SQL Server

有关详细信息，请参见我的博客中的文章：

• 分析函数：SUM，AVG，ROW_NUMBER-在中模拟MySQL。

这还将向您显示有多少重复项，并将不连接而对结果进行排序

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

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

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，它就像

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

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

最快的重复项删除查询过程：

/* 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;

我个人这个查询解决了我的问题：

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

该脚本的作用是在表中显示不止一次存在的所有订户ID，并找到重复的数量。

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    |                |

希望对您有帮助！

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

用表格替换城市。更换名称与字段名称

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

    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 |
+---------------------+------------+---+

要快速查看重复的行，您可以运行一个简单的查询

在这里，我要查询表并列出具有相同user_id，market_place和sku的所有重复行：

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，您必须确定要删除的行。例如ID较低的人（通常是年龄较大的人）或其他一些日期信息。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，请仔细检查是否删除了正确的记录。在这里，我从重复项中选择要删除的记录（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，我运行删除查询以删除重复对象：

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。

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。外部子查询只能返回一列，因为它用作运算符'= any'的操作数

Powerlord的答案的确是最好的，我建议再进行一次更改：使用LIMIT来确保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和进行联接，则使用LIMIT是一个好习惯。从较小的值开始，检查查询的强度，然后增加限制。