Question

我正在尝试计算此查询中每个授权号的总天数，但是我收到一条错误，指出聚合函数无法对具有聚合的表达式执行。我该如何解决这个问题？有几个授权号码，希望最短和最长日期能够在几天内提供准确的结果。

SELECT 
    CASE 
        WHEN DBO.FN_AGE(CL_DOB, AH_ADMDT) >= 18 THEN 'ADULT' 
        WHEN DBO.FN_AGE(CL_DOB, AH_ADMDT) < 18 THEN 'CHILD'  
        ELSE NULL 
    END As AgeGroup,
    COUNT(DISTINCT AH_AUTHNO) AS 'Total AFC Admissions',
    SUM(DATEDIFF(DD, MIN(AH_ADMDT), MAX(AH_DISDT))) AS 'Total AFC Days',
    SUM(AD_UNTAUTH) AS 'Total Units'
FROM 
    PCHAUHPF AS A
    LEFT JOIN
    PCHAUDPF AS P
        ON A.AH_RCDID = P.ADF_AUHID
    JOIN
    GTWAUDPF AS G
        ON G.ADP_AUDID = P.AD_RCDID
    JOIN
    PCHCLTPF AS C
        ON C.CL_RCDID = A.AHF_CLTID
    JOIN
    PCHXSPPF AS X
        ON X.XP_RCDID = P.ADF_XSPID
    JOIN
    PCHCPTPF AS T
        ON T.CR_RCDID = X.XPF_CPTID
WHERE 
    AH_ADMDT BETWEEN '10-1-2013' AND '09-30-2014'
    AND AD_PAYSTS = 'A'
    AND CR_CODE IN ('H2016', 'T1020')
GROUP BY
    CASE 
        WHEN DBO.FN_AGE(CL_DOB, AH_ADMDT) >= 18 THEN 'ADULT' 
        WHEN DBO.FN_AGE(CL_DOB, AH_ADMDT) < 18 THEN 'CHILD'  
        ELSE NULL 
    END

Answer 1

据推测，您想要的表达式是以下之一：

SUM(DATEDIFF(DAY, AH_ADMDT, AH_DISDT)) AS [Total AFC Days],

或：

DATEDIFF(DAY, MIN(AH_ADMDT), MAX(AH_DISDT)) AS [Total AFC Days],

关于样式问题，我建议您使用单引号仅表示字符串和日期常量（而不是列别名）。并且，在使用datediff()时使用完整日期部件名称。这样，你（或其他人）不必费力地记住MM是否持续数月或数分钟。

Answer 2

以下是评论中我的建议的一个简单示例：

WITH cte AS (
  SELECT 
     MyGroupCol
  ,  MIN(MyDateCol) AS MinDate
  ,  MAX(MyDateCol) AS MaxDate
  FROM MyTables
  WHERE SomeConditions=true
  GROUP BY MyGroupCol
)
SELECT 
  MyGroupCol
, SUM(DateDiff(dd, MinDate, MaxDate))
FROM cte
GROUP BY MyGroupCol

解决聚合中的聚合问题

2 个答案: