如果你想删除重复使用DISTINCT

否则使用这个查询:

SELECT users.*，COUNT(user_ID) as user FROM users GROUP BY user_name有用户>

对GROUP BY子句执行SELECT操作。假设name是你想要在其中找到重复项的列:

SELECT name, COUNT(*) c FROM table GROUP BY name HAVING c > 1;

这将返回一个在第一列中包含名称值的结果，以及该值在第二列中出现次数的计数。

我从下面开始改进:

SELECT 
    col, 
    COUNT(col)
FROM
    table_name
GROUP BY col
HAVING COUNT(col) > 1; 

SELECT ColumnA, COUNT( * )
FROM Table
GROUP BY ColumnA
HAVING COUNT( * ) > 1

作为利维克的答案的一个变体，它可以让你找到重复结果的id，我使用了以下方法:

SELECT * FROM table1 WHERE column1 IN (SELECT column1 AS duplicate_value FROM table1 GROUP BY column1 HAVING COUNT(*) > 1)

CREATE TABLE tbl_master
    (`id` int, `email` varchar(15));

INSERT INTO tbl_master
    (`id`, `email`) VALUES
    (1, 'test1@gmail.com'),
    (2, 'test2@gmail.com'),
    (3, 'test1@gmail.com'),
    (4, 'test2@gmail.com'),
    (5, 'test5@gmail.com');

QUERY : SELECT id, email FROM tbl_master
WHERE email IN (SELECT email FROM tbl_master GROUP BY email HAVING COUNT(id) > 1)