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
从表中的重复记录检查。
select * from users s
where rowid < any
(select rowid from users k where s.name = k.name and s.email = k.email);
or
select * from users s
where rowid not in
(select max(rowid) from users k where s.name = k.name and s.email = k.email);
删除表中的重复记录。
delete from users s
where rowid < any
(select rowid from users k where s.name = k.name and s.email = k.email);
or
delete from users s
where rowid not in
(select max(rowid) from users k where s.name = k.name and s.email = k.email);
其他回答
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,COUNT(email)
FROM users
WHERE email IN (
SELECT email
FROM users
GROUP BY email
HAVING COUNT(email) > 1)
派对晚了一点，但我找到了一个非常酷的解决方法来查找所有重复的ID：
SELECT email, GROUP_CONCAT(id)
FROM users
GROUP BY email
HAVING COUNT(email) > 1;
SELECT id，COUNT（id）FROM table1 GROUP BY id HAVING COUNT；
我认为这可以正确地搜索特定列中的重复值。
select name, email
, case
when ROW_NUMBER () over (partition by name, email order by name) > 1 then 'Yes'
else 'No'
end "duplicated ?"
from users
