使用一个字段很容易找到重复项:

SELECT email, COUNT(email) 
FROM users
GROUP BY email
HAVING COUNT(email) > 1

所以如果我们有一张桌子

ID   NAME   EMAIL
1    John   asd@asd.com
2    Sam    asd@asd.com
3    Tom    asd@asd.com
4    Bob    bob@asd.com
5    Tom    asd@asd.com

这个查询将告诉我们John、Sam、Tom和Tom,因为他们都有相同的电子邮件。

然而,我想要的是获得相同电子邮件和名称的副本。

也就是说,我想得到“汤姆”,“汤姆”。

我需要这个的原因是:我犯了一个错误,允许插入重复的名称和电子邮件值。现在我需要删除/更改重复项,所以我需要先找到它们。


当前回答

试试看:

                DECLARE @myTable TABLE
                (
                    id INT,
                    name VARCHAR(10),
                    email VARCHAR(50)
                );

                INSERT @myTable
                VALUES
                (1, 'John', 'John-email');
                INSERT @myTable
                VALUES
                (2, 'John', 'John-email');
                INSERT @myTable
                VALUES
                (3, 'fred', 'John-email');
                INSERT @myTable
                VALUES
                (4, 'fred', 'fred-email');
                INSERT @myTable
                VALUES
                (5, 'sam', 'sam-email');
                INSERT @myTable
                VALUES
                (6, 'sam', 'sam-email');


                WITH cte
                AS (SELECT ROW_NUMBER() OVER (ORDER BY (SELECT 1)) AS rowNum,
                           *
                    FROM @myTable)
                SELECT c1.id,
                       c1.name,
                       c1.email
                FROM cte AS c1
                WHERE 1 <
                (
                    SELECT COUNT(c2.rowNum)
                    FROM cte AS c2
                    WHERE c1.name = c2.name
                          AND c1.email = c2.email
                );

其他回答

表结构:

ID   NAME   EMAIL
1    John   asd@asd.com
2    Sam    asd@asd.com
3    Tom    asd@asd.com
4    Bob    bob@asd.com
5    Tom    asd@asd.com

解决方案1:

SELECT *,
       COUNT(*)
FROM users t1
INNER JOIN users t2
WHERE t1.id > t2.id
  AND t1.name = t2.name
  AND t1.email=t2.email

解决方案2:

SELECT name,
         email,
       COUNT(*)
FROM users
GROUP BY name,
         email
HAVING COUNT(*) > 1
 SELECT name, email 
    FROM users
    WHERE email in
    (SELECT email FROM users
    GROUP BY email 
    HAVING COUNT(*)>1)

您可以使用SELECT DISTINCT关键字来消除重复项。您还可以按名称进行筛选,并在表中获取具有该名称的所有人。

与其他答案不同,您可以查看包含所有列(如果有的话)的整个记录。在row_number函数的PARTITION BY部分中,选择所需的唯一/双工列。

SELECT  *
FROM    (
 SELECT a.*
 ,      Row_Number() OVER (PARTITION BY Name, Age ORDER BY Name) AS r
 FROM   Customers AS a
)       AS b
WHERE   r > 1;

当您想选择所有字段中的所有重复记录时,可以这样写

CREATE TABLE test (
        id      bigint GENERATED ALWAYS AS IDENTITY PRIMARY KEY
,       c1      integer
,       c2      text
,       d       date DEFAULT now()
,       v       text
);

INSERT INTO test (c1, c2, v) VALUES
(1, 'a', 'Select'),
(1, 'a', 'ALL'),
(1, 'a', 'multiple'),
(1, 'a', 'records'),
(2, 'b', 'in columns'),
(2, 'b', 'c1 and c2'),
(3, 'c', '.');
SELECT * FROM test ORDER BY 1;

SELECT  *
FROM    test
WHERE   (c1, c2) IN (
 SELECT c1, c2
 FROM   test
 GROUP  BY 1,2
 HAVING count(*) > 1
)
ORDER   BY 1;

在PostgreSQL中测试。

我们如何计算重复值??重复2次或大于2次。只是数数他们,而不是分组。

简单到

select COUNT(distinct col_01) from Table_01