mysql - 从类型 'DBNull' 到类型 'Double' 的转换无效

Question

在我的代码中，我理解查询获取空值并抛出此错误。但是由于我的查询有点复杂，我不明白如何检查空值并避免此错误。请帮助我更正此查询。

SELECT (SUM(charges) + SUM(behaviour) + SUM(admission) + SUM(properInformation) + SUM(hygine) + SUM(treatment))/(count(doctorID) * 6) AverageRating, COUNT(ID) RatingCount from ratings where doctorID = '" + doctorID + "'

Answer 1

如果您希望查询不返回NULL，您可以用IFNULL 包围表达式，将可能的NULL 转换为0，例如；

SELECT IFNULL((SUM(charges) + SUM(behaviour) + SUM(admission) + 
               SUM(properInformation) + SUM(hygine) + SUM(treatment))
              /(count(doctorID) * 6), 0) AverageRating, 
       COUNT(ID) RatingCount 
FROM ratings 
WHERE doctorID = '" + doctorID + "'