Question

我有以下MySQL表（images）：

+----+------------+-----------------------+
| id | gallery_id | path                                                                                                    |
+----+------------+-----------------------+
| 58 |       NULL | 58.jpg 
| 59 |       NULL | 59.jpg 
| 66 |          9 | 9-001.jpg 
| 67 |          9 | 9-002.jpg 
| 68 |         10 | 10-001.jpg 
| 69 |         10 | 10-002.jpg 
...

我想选择gallery_id为null的行，或者如果它不为null，则按gallery_id对其进行分组。所以预期的结果是：

+----+------------+-----------------------+
| id | gallery_id | path                                                                                                    |
+----+------------+-----------------------+
| 58 |       NULL | 58.jpg 
| 59 |       NULL | 59.jpg 
| 66 |          9 | 9-001.jpg 
| 68 |         10 | 10-001.jpg 
...

我尝试使用coalesce函数来实现此结果：

select * from `images` group by coalesce(gallery_id, id);

但它只返回gallery_id为空的行。请告诉我，我做错了什么？提前谢谢。

Answer 1

你可以使用with in in subselect和group by

select * from `images`
where (gallery_id, id ) in (select gallery_id, min(id) 
                             from `images`
                             where gallery_id is not null
                             group by gallery_id)

Answer 2

SELECT * FROM `images` where gallery_id is not null group by gallery_id 
UNION
SELECT * FROM `images` where gallery_id is null

Answer 3

您的请求没有意义。 GROUP BY SELECT * NULL是非感性的，因为不清楚大多数列值的来源。

您似乎想要确定优先顺序。。。每个select i.* from images i where i.gallery_id is null or i.id in (select min(i2.id) from images i2 group by i2.gallery_id);一行，然后每个库一个。

我建议：

use

作为奖励，这是标准的SQL，应该适用于任何数据库。

coalesce的SQL组不按预期工作

3 个答案: