3

我想尝试使用外连接来找出表中的重复项：

如果一个表有主索引 ID，那么下面的外连接可以找出重复的名称：

mysql> select * from gifts;
+--------+------------+-----------------+---------------------+
| giftID | name       | filename        | effectiveTime       |
+--------+------------+-----------------+---------------------+
|      2 | teddy bear | bear.jpg        | 2010-04-24 04:36:03 |
|      3 | coffee     | coffee123.jpg   | 2010-04-24 05:10:43 |
|      6 | beer       | beer_glass.png  | 2010-04-24 05:18:12 |
|     10 | heart      | heart_shape.jpg | 2010-04-24 05:11:29 |
|     11 | ice tea    | icetea.jpg      | 2010-04-24 05:19:53 |
|     12 | cash       | cash.png        | 2010-04-24 05:27:44 |
|     13 | chocolate  | choco.jpg       | 2010-04-25 04:04:31 |
|     14 | coffee     | latte.jpg       | 2010-04-27 05:49:52 |
|     15 | coffee     | espresso.jpg    | 2010-04-27 06:03:03 |
+--------+------------+-----------------+---------------------+
9 rows in set (0.00 sec)

mysql> select * from gifts g1 LEFT JOIN (select * from gifts group by name) g2 
         on g1.giftID = g2.giftID where g2.giftID IS NULL;
+--------+--------+--------------+---------------------+--------+------+----------+---------------+
| giftID | name   | filename     | effectiveTime       | giftID | name | filename | effectiveTime |
+--------+--------+--------------+---------------------+--------+------+----------+---------------+
|     14 | coffee | latte.jpg    | 2010-04-27 05:49:52 |   NULL | NULL | NULL     | NULL          |
|     15 | coffee | espresso.jpg | 2010-04-27 06:03:03 |   NULL | NULL | NULL     | NULL          |
+--------+--------+--------------+---------------------+--------+------+----------+---------------+
2 rows in set (0.00 sec)

但是如果表没有主索引 ID，那么外连接仍然可以用来找出重复项吗？

PS 也感谢任何非外部连接解决方​​案。如果可能，我想检查在没有主 ID 索引时是否可以使用外连接来完成。谢谢你的帮助。

4

2 回答 2

2

使用EXISTS条款：

SELECT  *
FROM    gifts go
WHERE   EXISTS
        (
        SELECT  NULL
        FROM    gifts gi
        WHERE   gi.name = go.name
        LIMIT 1, 1
        )
于 2010-04-27T15:02:31.127 回答
2

您可以使用分组来找出重复值：

select name
from gifts
group by name
having count(*) > 1
于 2010-04-27T15:06:01.663 回答