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
select id,name,COUNT(*) from user group by Id,Name having COUNT(*)>1
其他回答
试试看：
SELECT name, email
FROM users
GROUP BY name, email
HAVING ( COUNT(*) > 1 )
这里最重要的是要有最快的功能。还应确定副本索引。自连接是一个很好的选项，但为了获得更快的功能，最好先查找具有重复行的行，然后与原始表连接以查找重复行的id。最后按除id之外的任何列排序，以使重复的行彼此靠近。
SELECT u.*
FROM users AS u
JOIN (SELECT username, email
FROM users
GROUP BY username, email
HAVING COUNT(*)>1) AS w
ON u.username=w.username AND u.email=w.email
ORDER BY u.email;
在使用Microsoft Access的情况下，此方法有效：
CREATE TABLE users (id int, name varchar(10), email varchar(50));
INSERT INTO users VALUES (1, 'John', 'asd@asd.com');
INSERT INTO users VALUES (2, 'Sam', 'asd@asd.com');
INSERT INTO users VALUES (3, 'Tom', 'asd@asd.com');
INSERT INTO users VALUES (4, 'Bob', 'bob@asd.com');
INSERT INTO users VALUES (5, 'Tom', 'asd@asd.com');
SELECT name, email, COUNT(*) AS CountOf
FROM users
GROUP BY name, email
HAVING COUNT(*)>1;
DELETE *
FROM users
WHERE id IN (
SELECT u1.id
FROM users u1, users u2
WHERE u1.name = u2.name AND u1.email = u2.email AND u1.id > u2.id
);
感谢Tancrede Chazallet的删除代码。
select name, email
, case
when ROW_NUMBER () over (partition by name, email order by name) > 1 then 'Yes'
else 'No'
end "duplicated ?"
from users
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
);
