从连接表中获取孩子的总记录

Question

我使用的是 MariaDB 5.5，但对于此解决方案，它与 MySQL 相同。我有两个表，第一个包含画廊，第二个包含有关每个画廊中文件的信息。这是gallery 表的示例：

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

这是gallery_items表的示例：

+----+------+------------+-----+
| 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;

嗯，我不太擅长数据库，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

您需要GROUP BY 才能让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

Answer 2

说明您可以使用以下查询

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