如何在 Select 语句中使用聚合答案

【问题标题】：How to use Aggregate in Select Statement如何在 Select 语句中使用聚合
【发布时间】：2017-05-03 11:57:46
【问题描述】：

我的代码有问题，它似乎不接受我的选择查询中的聚合函数。我想用我的DATEDIFF(MONTH,MIN(LoanDateStart),MAX(LoanPaymentDue)) 实现的是我想获得总月数，然后使用月数来计算其余查询。

我收到了这个错误：

消息 130，级别 15，状态 1，第 11 行无法对包含聚合或子查询的表达式执行聚合函数。

有没有其他方法可以实现？

查询

SELECT 
ISNULL(SUM((CAST(((((lt.InterestRate/100) * lc.LoanAmount) +
 lc.LoanAmount) / ((dbo.fnNumberOfYears(CONVERT(VARCHAR(15), LoanDateStart, 
101), CONVERT(VARCHAR(15), LoanPaymentDue, 101)) * DATEDIFF(MONTH, 
MIN(LoanDateStart),  MAX(LoanPaymentDue)))  * 2)) AS DECIMAL(18,2)))),0)
FROM LoanContract lc 
INNER JOIN LoanType lt ON lt.LoanTypeID = lc.LoanTypeID 
WHERE lc.LoanTypeID = 1 AND lc.EmployeeID = 5

【问题讨论】：

标签： sql sql-server

【解决方案1】：

请尝试以下...

SELECT ISNULL( SUM( ( CAST( ( ( ( ( lt.InterestRate / 100 ) *
                                  LoanContract.LoanAmount ) +
                                LoanContract.LoanAmount ) /
                              ( ( dbo.fnNumberOfYears( CONVERT( VARCHAR( 15 ),
                                                                LoanDateStart,
                                                                101 ),
                                                       CONVERT( VARCHAR( 15 ),
                                                                LoanPaymentDue,
                                                                101 ) ) *
                                  monthsDifference ) *
                                2 ) ) AS DECIMAL( 18,
                                                  2 ) ) ) ),
               0 )
FROM ( SELECT LoanContractID AS LoanContractID,
              DATEDIFF( MONTH,
                        MIN( LoanDateStart ),
                        MAX( LoanPaymentDue ) ) AS monthsDifference
       FROM LoanContract
       GROUP BY LoanContractID
     ) AS monthsDifferenceFinder
INNER JOIN LoanContract ON LoanContract.LoanContractID = monthsDifferenceFinder.LoanContractID
INNER JOIN LoanType lt ON lt.LoanTypeID = LoanContract.LoanTypeID
WHERE lc.LoanTypeID = 1
  AND lc.EmployeeID = 5

请注意，我使用 LoanContractID 代替了 LoanContract 的主键，因为您没有说明那是什么。

问题的原因是 SUM()（聚合函数）对 MIN() 和 MAX() 的结果进行运算，它们本身就是聚合函数。这很困惑SQL-Server。

我已经通过编写一个子查询来解决这个问题，该子查询根据 LoanContract 的唯一标识符而不是每条记录的主键来确定每个 LoanContract 的两个值之间的差异。（您的数据是 3NF（第三范式）吗？）

然后基于LoanContractID 将子查询的结果连接到LoanContract，有效地将每个LoanContract 的monthsDifference 值附加到该LoanContract 的每条记录中。

SUM() 然后能够处理检索到的值，而不必尝试聚合正确聚合的值。

如果您有任何问题或cmets，请随时发表相应的评论。

【讨论】：

在答案中添加了说明。
感谢您的精彩解释，我现在完全明白为什么了。非常感谢@toonice

【解决方案2】：

无论如何，我现在解决了这个问题，我正在使用LEFT OUTER JOIN。

这是我的代码：

SELECT
ISNULL(SUM((CAST(((((lt.InterestRate/100) * lc.LoanAmount) + lc.LoanAmount) / ((dbo.fnNumberOfYears(CONVERT(VARCHAR(15), LoanDateStart, 101), CONVERT(VARCHAR(15), LoanPaymentDue, 101)) * x.NumberOfMonths)  * 2)) AS DECIMAL(18,2)))),0)
FROM LoanContract lc 
INNER JOIN LoanType lt ON lt.LoanTypeID = lc.LoanTypeID
LEFT OUTER JOIN(SELECT
lc.LoanTypeID AS 'LoanTypeID',
lc.EmployeeID AS 'EmployeeID', 
(DATEDIFF(MONTH, MIN(LoanDateStart),  MAX(LoanPaymentDue))) AS 'NumberOfMonths'
FROM LoanContract lc 
INNER JOIN LoanType lt ON lt.LoanTypeID = lc.LoanTypeID
GROUP BY lc.EmployeeID, lc.LoanTypeID) x ON x.EmployeeID = lc.EmployeeID AND x.LoanTypeID = lc.LoanTypeID 
WHERE lc.LoanTypeID = 1 AND lc.EmployeeID = 5

【讨论】：