我有 3 个表,如下所示:
mysql> select * from Raccoon;
+----+------------------+----------------------------------------------------------------------------------------------------+
| id | name | image_url |
+----+------------------+----------------------------------------------------------------------------------------------------+
| 3 | Jesse Coon James | http://www.pawfun.com/wp/wp-content/uploads/2010/01/rabbid.png |
| 4 | Bobby Coon | https://c2.staticflickr.com/6/5242/5370143072_dee60d0ce2_n.jpg |
| 5 | Doc Raccoon | http://images.encyclopedia.com/utility/image.aspx?id=2801690&imagetype=Manual&height=300&width=300 |
| 6 | Eddie the Rac | http://www.felid.org/jpg/EDDIE%20THE%20RACCOON.jpg |
+----+------------------+----------------------------------------------------------------------------------------------------+
4 rows in set (0.00 sec)
mysql> select * from Review;
+----+------------+-------------+---------------------------------------------+--------+
| id | raccoon_id | reviewer_id | review | rating |
+----+------------+-------------+---------------------------------------------+--------+
| 1 | 3 | 1 | This raccoon was a fine raccoon indeed. | 5 |
| 2 | 5 | 2 | This raccoon did not do much for me at all. | 2 |
| 3 | 3 | 1 | asdfsadfsadf | 5 |
| 4 | 5 | 2 | asdfsadf | 1 |
+----+------------+-------------+---------------------------------------------+--------+
4 rows in set (0.00 sec)
mysql> select * from Reviewer;
+----+---------------+
| id | reviewer_name |
+----+---------------+
| 1 | Kane Charles |
| 2 | Cameron Foale |
+----+---------------+
2 rows in set (0.00 sec)
我正在尝试构建一个选择查询,该查询将返回 Raccoon
中的所有列以及一个获取平均 Review.rating
的额外列(按 id 分组)。我面临的问题是,不能保证每个 Raccoon 的 Review
表中都会有行(由 FK 确定,raccoon_id
引用 Raccoon.id
。在 Review
表中存在零行的情况下(对于给定的 Raccoon.id,即 Review.raccoon_id),我希望查询返回 0
作为该浣熊的平均值。
下面是我正在使用的当前查询:
mysql> SELECT *, (SELECT IFNULL(AVG(rating),0) FROM Review WHERE raccoon_id=Raccoon.id GROUP BY raccoon_id) AS "AVG" FROM Raccoon ORDER BY "AVG" ASC;
+----+------------------+----------------------------------------------------------------------------------------------------+--------+
| id | name | image_url | AVG |
+----+------------------+----------------------------------------------------------------------------------------------------+--------+
| 3 | Jesse Coon James | http://www.pawfun.com/wp/wp-content/uploads/2010/01/rabbid.png | 5.0000 |
| 4 | Bobby Coon | https://c2.staticflickr.com/6/5242/5370143072_dee60d0ce2_n.jpg | NULL |
| 5 | Doc Raccoon | http://images.encyclopedia.com/utility/image.aspx?id=2801690&imagetype=Manual&height=300&width=300 | 1.5000 |
| 6 | Eddie the Rac | http://www.felid.org/jpg/EDDIE%20THE%20RACCOON.jpg | NULL |
+----+------------------+----------------------------------------------------------------------------------------------------+--------+
4 rows in set (0.00 sec)
正如您在上面看到的,对于 ID 为 4 和 6 的 Raccoon,查询不会返回 0
,它只是返回 NULL
。我需要它返回如下内容(注意顺序,首先按最低平均评论排序):
+----+------------------+----------------------------------------------------------------------------------------------------+--------+
| id | name | image_url | AVG |
+----+------------------+----------------------------------------------------------------------------------------------------+--------+
| 4 | Bobby Coon | https://c2.staticflickr.com/6/5242/5370143072_dee60d0ce2_n.jpg | 0.0000 |
| 6 | Eddie the Rac | http://www.felid.org/jpg/EDDIE%20THE%20RACCOON.jpg | 0.0000 |
| 5 | Doc Raccoon | http://images.encyclopedia.com/utility/image.aspx?id=2801690&imagetype=Manual&height=300&width=300 | 1.5000 |
| 3 | Jesse Coon James | http://www.pawfun.com/wp/wp-content/uploads/2010/01/rabbid.png | 5.0000 |
+----+------------------+----------------------------------------------------------------------------------------------------+--------+
最佳答案
在你的子查询之外使用 IFNULL
因为它会返回 null 如果在外表上没有匹配,
IFNULL((SELECT AVG(rating) FROM Review WHERE raccoon_id=Raccoon.id GROUP BY raccoon_id), 0) AS "AVG"
或者你也可以使用LEFT JOIN
,
SELECT ra.id, ra.name, ra.image_url,
IFNULL(AVG(rv.rating),0)AS "AVG"
FROM Raccoon ra
LEFT JOIN Review rv
ON rv.raccoon_id = ra.id
GROUP BY ra.id, ra.name, ra.image_url
ORDER BY "AVG" ASC;
关于MySQL AVG() 如果为 NULL 则返回 0,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33990429/