从连接表中获取子项的总记录

Question

我正在使用MariaDB 5.5,但对于这个解决方案,它将与MySQL相同.我有两个表,第一个包含画廊,第二个包含有关每个库中文件的信息.这是表的示例gallery:

+----+-------+-----+
| id | name  | ... |
+----+-------+-----+
| 1  | test1 | ... |
| 2  | test2 | ... |
| 3  | test3 | ... |
| 4  | test4 | ... |
+----+-------+-----+

这是表的示例gallery_items:

+----+------+------------+-----+
| id | file | gallery_id | ... |
+----+------+------------+-----+
| 1  | img1 | 3          | ... |
| 2  | img2 | 2          | ... |
| 3  | img3 | 2          | ... |
| 4  | img4 | 1          | ... |
+----+------+------------+-----+

所以我尝试了这段代码:

SELECT gallery.*, COUNT(gallery_items.id) AS items FROM gallery JOIN gallery_items WHERE gallery_items.gallery_id = gallery.id;

好吧,我对数据库不太满意,所以这就是我寻求帮助的原因.这是我的预期结果:

+----+-------+-------+-----+
| id | name  | items | ... |
+----+-------+-------+-----+
| 1  | test1 | 1     | ... |
| 2  | test2 | 2     | ... |
| 3  | test3 | 1     | ... |
| 4  | test4 | 0     | ... |
+----+-------+-------+-----+

Answer 1

你需要GROUP BY为了COUNT工作

SELECT gallery.*, COUNT(gallery_items.id) AS items FROM gallery 
LEFT JOIN gallery_items ON gallery_items.gallery_id = gallery.id
GROUP BY gallery.id, gallery.name