Question

我正在编写我认为对于我的医生客户来说相对简单的Access DB，他的实践非常少。我在报告患者的平衡方面遇到了麻烦。

我有3张桌子：

Patients (id, [First Name], [Last Name], {other cols})
Treatments (id, Fee, Patient_id, {other cols})
Payments (id, Amount, Patient_id, {other cols})

我想要一个查询，只是告诉我治疗费用总额，支付总金额和当前余额。我写了以下内容......

Select 
 Patients.[Last Name],
 Patients.[First Name], 
 SUM(select SUM(Treatments.Fee) from Treatments Where Treatments.Patient=@PatientID)
AS CHARGES, 
 SUM(select SUM(PAYMENTS.AMOUNT) from PAYMENTS Where PAYMENTS.Patient=@PatientID)
AS PAYMENTS, 
SUM(
 (select SUM(Treatments.Fee) from Treatments Where Treatments.Patient=@PatientID)
- (select SUM(PAYMENTS.AMOUNT) from PAYMENTS Where PAYMENTS.Patient=@PatientID)
)
as Balance
FROM Patients,Treatments,PAYMENTS
WHERE Patients.ID = @PatientID
GROUP BY Patients.[Last Name],Patients.[First Name]

费用和付款列工作正常，但我给出的余额很奇怪。以下是基于：

该患者的5个治疗项目，金额为50,25,35,45,125（280）。
3笔付款条目，金额为15,60,25（100）。

我希望：费用= 280美元，付款= 100美元，余额= 180美元

但我明白了：费用= 280美元，付款= 100美元，余额= 2,700美元

Answer 1

Select 
 Patients.[Last Name],
 Patients.[First Name], 
 SUM( treatments.Fee) 
AS CHARGES, 
 SUM(PAYMENTS.AMOUNT) 
AS PAYMENTS, 
SUM(Treatments.Fee) 
- SUM(PAYMENTS.AMOUNT) 
as Balance
FROM Patients a join Treatments b on a.id=b.patientid join PAYMENTS c on a.id=c.patientid
WHERE Patients.ID = @PatientID
GROUP BY Patients.[Last Name],Patients.[First Name]

Answer 2

使用此FROM子句：

FROM Patients,Treatments,PAYMENTS

您正在创建笛卡尔积。这意味着您的余额计算将乘以每个表中的记录数（患者为1，治疗为5，付款为3）：

15 *费用总额 - 15 *付款总额= 15 * 280-15 * 100 = 4200-1500 = 2700

在您的子查询中进行计算时，您的FROM子句中不需要处理和付款。请改用此语句：

Select 
 Patients.[Last Name],
 Patients.[First Name], 
 SUM(select SUM(Treatments.Fee) from Treatments Where Treatments.Patient=@PatientID)
AS CHARGES, 
 SUM(select SUM(PAYMENTS.AMOUNT) from PAYMENTS Where PAYMENTS.Patient=@PatientID)
AS PAYMENTS, 
SUM(
 (select SUM(Treatments.Fee) from Treatments Where Treatments.Patient=@PatientID)
- (select SUM(PAYMENTS.AMOUNT) from PAYMENTS Where PAYMENTS.Patient=@PatientID)
)
as Balance
FROM Patients
WHERE Patients.ID = @PatientID
GROUP BY Patients.[Last Name],Patients.[First Name]

使用减法访问SQL SUM

2 个答案: