DB2 中的“无效使用聚合函数或 OLAP 函数”错误

Question

我在下面附上了我的代码供您查看。我的错误来自第四个 CASE 语句，我使用数学函数来计算信息。有谁知道是什么导致了错误以及如何解决它？我猜问题出在我最后的 ORDER BY 函数中。

SELECT
H.DATE AS "DATE"
,TRIM(H.NUMBER) AS "NUMBER"
,CASE WHEN TT.RY_DATE IS NULL THEN '' ELSE CHAR(TT.RY_DATE) END AS "RY DATE"
,CASE WHEN T.PT_DATE = '0001-01-01' THEN 'N' ELSE 'Y' END AS "PT"
,T.ON AS "ON"
,CASE WHEN H.CLASS = '0.00' THEN H.CLASS ELSE H.RATED END AS "CLASS"
,SUBSTR(TRIM(S.NAME), 1, LENGTH(TRIM(S.NAME))-2) AS "CITY"
,H.STATE AS "STATE"
,H.ZIPCODE AS "ZIP"
,SUBSTR(TRIM(CN.NAME), 1, LENGTH(TRIM(CN.NAME))-2) AS "DCITY"
,H.STATE AS "DSTATE"
,H.ZIPCODE AS "DZIP"
,H.WGT AS "WEIGHT"
,CASE 
WHEN Q.AMOUNT IS NULL THEN (Y.CHGS - Z.AMOUNT)
WHEN Q.AMOUNT IS NOT NULL THEN (Y.CHGS - (SUM(Q.AMOUNT +  Z.AMOUNT))) ELSE ''
END AS "LL"
,Z.AMOUNT AS "FU"
,SUM(Q.AMOUNT) AS "AC"
,Y.CHGS AS "CHARGES"

FROM A.COST H

INNER JOIN A.MASTER S
ON H.CITY = S.CITY

INNER JOIN A.MASTER CN
ON H.CITY = CN.CITY

INNER JOIN A.SPEC Z
ON Z.NUMBER = H.NUMBER
AND Z.DATE = H.DATE
AND Z.TYPE = 'F'

INNER JOIN A.ALT Y
ON Y.NUMBER = H.NUMBER
AND Y.DATE = H.DATE

LEFT OUTER JOIN A.SPEC. Q
ON Q.NUMBER = H.NUMBER
AND Q.DATE = H.DATE
AND Q.TYPE = 'S'

LEFT OUTER JOIN A.T TT
ON H.NUMBER = TT.NUMBER

LEFT OUTER JOIN A.TIME T
ON T.NUMBER = H.NUMBER

WHERE H.CTRL = '000000'
AND (MONTH(CURRENT DATE)-1) = MONTH(H.DATE)
AND H.DATE > CURRENT DATE - 90 DAYS

GROUP BY
H.DATE 
,TRIM(H.NUMBER)
,CASE WHEN TT.D_DATE IS NULL THEN '' ELSE CHAR(TT.D_DATE) END
,CASE WHEN T.PT_DATE = '0001-01-01' THEN 'N' ELSE 'Y' END
,T.ON
,CASE WHEN H.CLASS = '0.00' THEN H.CLASS ELSE H.RATED END
,SUBSTR(TRIM(S.NAME), 1, LENGTH(TRIM(S.NAME))-2)
,H.STATE
,H.ZIPCODE 
,SUBSTR(TRIM(CN.NAME), 1, LENGTH(TRIM(CN.NAME))-2)
,H.STATE 
,H.ZIPCODE 
,H.WGT 
,CASE WHEN Q.AMOUNT IS NULL THEN (Y.CHGS - Z.AMOUNT)
WHEN Q.AMOUNT IS NOT NULL THEN (Y.CHGS - (SUM(Q.AMOUNT + Z.AMOUNT))) END
,Z.AMOUNT 
,Y.CHGS

ORDER BY DATE

Answer 1

问题是您试图在 SELECT 中同时引用单个和聚合的 Q.AMOUNT 值。

如果我正确理解您的意图，您可能打算在您的第四个 CASE 中这样做：

Y.CHGS - (COALESCE(SUM(Q.AMOUNT), 0) +  Z.AMOUNT) AS "LL"

如果组中的每个Q.AMOUNT 都为NULL，则SUM(Q.AMOUNT) 也将为NULL 并替换为零，这实质上将使表达式等效于Y.CHGS - Z.AMOUNT。

您无需在 GROUP BY 中重复此表达式，因为其中的所有引用都已经是 GROUP BY 条件或正在聚合。

COALESCE 函数接受多个参数并返回第一个不为 NULL 的参数。

Answer 2

在第四种情况下，使用额外的SUM 函数是没有用的。尝试这样的事情：-

CASE 
WHEN Q.AMOUNT IS NULL THEN (Y.CHGS - Z.AMOUNT)
WHEN Q.AMOUNT IS NOT NULL THEN (Y.CHGS - Q.AMOUNT +  Z.AMOUNT) 
ELSE ''
END AS "LL"

希望对你有帮助。

Answer 3

从 GROUP BY 中删除以下内容：

CASE WHEN Q.AMOUNT IS NULL THEN (Y.CHGS - Z.AMOUNT)
WHEN Q.AMOUNT IS NOT NULL THEN (Y.CHGS - (SUM(Q.AMOUNT + Z.AMOUNT))) END