MySQL AVG() 如果为 NULL，则返回 0答案

【问题标题】：MySQL AVG() return 0 if NULLMySQL AVG() 如果为 NULL，则返回 0
【发布时间】：2015-11-30 02:01:18
【问题描述】：

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额外的列，该列获取平均 Review.rating（按 id 分组）。我面临的问题是，不能保证Review 表中每个 Raccoon 都会存在行（由 FK 确定，raccoon_id 引用Raccoon.id。在存在零行的情况下在 Review 表中（对于给定的 Raccoon.id，即 Review.raccoon_id），我希望查询返回 0 作为该 Raccoon 的平均值。

以下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 Raccoons，查询没有返回 0，它只是返回 NULL。我需要它返回如下内容（注意排序，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 |
+----+------------------+----------------------------------------------------------------------------------------------------+--------+

【问题讨论】：

标签： mysql sql if-statement null average

【解决方案1】：

在子查询之外使用IFNULL，因为如果外部表不匹配，它将返回 null，

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;

【讨论】：

【解决方案2】：

您不想在子查询中使用group by。这很危险，因为它可能返回多行（尽管where 阻止了这种情况）。更重要的是，没有group by，子查询是一个总是返回一行的聚合查询。所以，你可以把逻辑放在子查询中：

SELECT r.*,
       (SELECT COALESCE(AVG(rev.rating),0)
        FROM Review rev
        WHERE rev.raccoon_id = r.id
       ) AS "AVG"
FROM Raccoon r
ORDER BY "AVG" ASC;

另外：当您有相关子查询时，请始终使用限定列名。这是防止将来出现问题的好习惯。

【讨论】：

【解决方案3】：

您可以使用内置函数 COALESCE() 尝试以下 SQL 语句：

SELECT *,  COALESCE((SELECT AVG(rating) FROM Review WHERE raccoon_id=Raccoon.id GROUP BY raccoon_id), 0) AS "AVG" FROM Raccoon ORDER BY "AVG" ASC;

你可以找到这个函数的使用手册here。

如果你更喜欢使用 IFNULL，你可以使用

SELECT *, IFNULL((SELECT AVG(rating) FROM Review WHERE raccoon_id=Raccoon.id GROUP BY raccoon_id), 0) AS "AVG" FROM Raccoon ORDER BY "AVG" ASC;

您似乎误解了函数的范围。

但是，我认为更好的方法是使用左外连接，而不是子查询，这是我写的查询供您参考：

select Raccoon.id, Raccoon.name, Raccoon.image_url, IFNULL(AVG(rating),0) avg from Raccoon LEFT OUTER join Review ON raccoon_id =Raccoon.id GROUP BY raccoon.id ORDER BY AVG ASC;

然后你会得到以下结果：

（对不起，我不知道如何发布此查询的输出，然后我截取了屏幕截图 :-)

希望这会有所帮助。

【讨论】：