MySQL 2 列不工作的百分比 =NULL

Question

我在MYSQL中有如下语句：

SELECT Site, Areateam, 
SUM( IF( year = '15-16', 1, 0 ) ) "Y2", 
SUM( IF( year = '14-15', 1, 0 ) ) "Y1",
SUM('Y2') / SUM('Y1')* 100 AS Diff
FROM CD2015_EmailIncidents
WHERE Areateam = 'Greater Manchester'
GROUP BY Site

返回以下结果：

**Site           |Areateam             |Y2  |Y1  |Diff**

Acute Trust      |Greater Manchester   |0   |1   |NULL

Care Home        |Greater Manchester   |3   |22  |NULL

CD Store Room    |Greater Manchester   |7   |4   |NULL

College Greater  |Greater Manchester   |0   |1   |NULL

我似乎无法用正确的语法来显示 Y2 和 Y1 之间的百分比差异，因为它一直显示为 NULL

非常感谢 最大

Answer 1

您的问题是您正在对字符串值进行算术运算：

SUM('Y2') / SUM('Y1')* 100 AS Diff

MySQL 进行静默转换，因此它将值转换为数字。没有前导数字，因此它们被转换为 0。

MySQL 然后为0/0 返回NULL。

Juergen 对此有正确的解决方案。您的整体查询可以表示为：

SELECT Site, Areateam, 
       SUM(year = '15-16') as Y2, 
       SUM(year = '14-15') as Y1,
       100 * SUM(year = '15-16') / SUM(year = '14-15') AS Diff
FROM CD2015_EmailIncidents
WHERE Areateam = 'Greater Manchester'
GROUP BY Site;

Answer 2

您不能在select 子句中使用计算的列名。并首先乘以 100 以避免整数限制

SUM(year = '15-16') * 100 / SUM(year = '14-15')  AS Diff

Answer 3

那是因为“Y1”和“Y2”在该上下文中不存在。

SELECT Site, Areateam, 
    SUM( IF( year = '15-16', 1, 0 ) ) "Y2", 
    SUM( IF( year = '14-15', 1, 0 ) ) "Y1",
    SUM( IF( year = '15-16', 1, 0 ) ) * 100 / 
    SUM( IF( year = '14-15', 1, 0 ) AS Diff
FROM CD2015_EmailIncidents
WHERE Areateam = 'Greater Manchester'
GROUP BY Site

在此处检查舍入ROUND OR TRUNC