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
你可能想试试这个
SELECT NAME, EMAIL, COUNT(*)
FROM USERS
GROUP BY 1,2
HAVING COUNT(*) > 1
其他回答
SELECT name, email,COUNT(email)
FROM users
WHERE email IN (
SELECT email
FROM users
GROUP BY email
HAVING COUNT(email) > 1)
通过使用CTE，我们也可以找到类似这样的重复值
with MyCTE
as
(
select Name,EmailId,ROW_NUMBER() over(PARTITION BY EmailId order by id) as Duplicate from [Employees]
)
select * from MyCTE where Duplicate>1
select id,name,COUNT(*) from user group by Id,Name having COUNT(*)>1
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
);
你可能想试试这个
SELECT NAME, EMAIL, COUNT(*)
FROM USERS
GROUP BY 1,2
HAVING COUNT(*) > 1
