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
                );

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

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
, case 
when ROW_NUMBER () over (partition by name, email order by name) > 1 then 'Yes'
else 'No'
end "duplicated ?"
from users

这个问题在上面的所有答案中都得到了很好的回答。但我想列出所有可能的方式，我们可以通过各种方式来做到这一点，这可能会让我们了解如何做到，寻求者可以选择最适合他/她的需求的解决方案，因为这是SQL开发人员遇到不同业务用例或在访谈中遇到的最常见的查询之一。

创建示例数据

我将仅从这个问题中设置一些示例数据开始。

Create table NewTable (id int, name varchar(10), email varchar(50))
INSERT  NewTable VALUES (1,'John','asd@asd.com')
INSERT  NewTable VALUES (2,'Sam','asd@asd.com')
INSERT  NewTable VALUES (3,'Tom','asd@asd.com')
INSERT  NewTable VALUES (4,'Bob','bob@asd.com')
INSERT  NewTable VALUES (5,'Tom','asd@asd.com')

1.使用groupby子句

SELECT
    name,email, COUNT(*) AS Occurence
    FROM NewTable
    GROUP BY name,email
    HAVING COUNT(*)>1

工作原理：

GROUP BY子句按中的值将行分组姓名和电子邮件栏。然后，COUNT（）函数返回数字每个组的出现次数（姓名、电子邮件）。然后，HAVING子句保持仅重复组，这些组包含多个发生

2.使用CTE：

要返回每个重复行的整个行，请使用公共表表达式（CTE）将上述查询的结果与NewTable表连接：

WITH cte AS (
    SELECT
        name, 
        email, 
        COUNT(*) occurrences
    FROM NewTable
    GROUP BY 
        name, 
        email
    HAVING COUNT(*) > 1
)
SELECT 
    t1.Id,
    t1.name, 
    t1.email
FROM  NewTable t1
    INNER JOIN cte ON 
        cte.name = t1.name AND 
        cte.email = t1.email
ORDER BY 
    t1.name, 
    t1.email;

3.使用ROW_NUMBER（）函数

WITH cte AS (
    SELECT 
        name, 
        email, 
        ROW_NUMBER() OVER (
            PARTITION BY name,email
            ORDER BY name,email) rownum
    FROM 
        NewTable t1
) 
SELECT 
  * 
FROM 
    cte 
WHERE 
    rownum > 1;

工作原理：

ROW_NUMBER（）将NewTable表的行按名称和电子邮件列中的值分配到分区中。重复的行在名称和电子邮件列中具有重复的值，但行号不同外部查询删除每个组中的第一行。

好吧，现在我相信，你可以有正确的想法，如何找到重复，并应用逻辑在所有可能的场景中找到重复。谢谢

我想这对你有帮助

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