Question

我正在使用PHP，我有以下两个表：

REVIEWS

|  id  |  Item   | editor_rating | user_votes  |
------------------------------------------------
|   1  | item_1  |      18.1     |      415    |
|   2  | item_2  |      17.1     |      371    |
|   3  | item_3  |      14.7     |      111    |

|   8  | item_8  |      15.3     |      215    |
|   9  | item_9  |      17.7     |      119    |
|  10  | item_10 |      17.0     |      66     |


FILTERS

|  id  |  Item   |   published   |
----------------------------------
|   1  | item_1  |       1       |
|   2  | item_2  |       1       |
|   3  | item_3  |       1       |

|   8  | item_8  |       1       |
|   9  | item_9  |       1       |
|  10  | item_10 |       1       |

以下查询正常工作并返回所有行：

SELECT * FROM reviews AS r 
JOIN filters AS f ON (r.id = f.id) 
WHERE f.published = 1

所以我的方式正确。现在我想对记录做一些数学运算。例如 sum 字段 editor_rating 和 user_votes ，并列出DESC排序的行。我尝试了以下但它返回不正确的结果（不是MySQL错误，总和错误）：

 SELECT *, r.editor_rating+r.user_votes AS total FROM reviews AS r 
 JOIN filters AS f ON (r.id = f.id) 
 WHERE f.published = 1 ORDER BY total DESC

上表包含实际的dB数据。通过总结 editor_rating 和 user_votes ，我应该按如下方式订购一个列表：

18.1 + 415 = 433.1 = Item_1
17.1 + 371 = 388.1 = Item_2
15.3 + 215 = 230.3 = Item_8

而是我得到以下列表：

Item_8
Item_9
Item_10

Answer 1

首先检查这些值是否为varchar，然后必须先将其强制转换：

 SELECT *, cast(r.editor_rating as unsigned )+ cast(r.user_rating as unsigned ) AS total FROM reviews AS r JOIN filters AS f ON (r.id = f.id) WHERE f.published = 1 ORDER BY total DESC

当我首先将它转换为无符号整数时，它会为我运行..

多个表上的数学运算

1 个答案: