Question

我有一个用于表示时间的smallint列。我已经编写了以下代码来根据时间推导出结果：

        CASE
    WHEN 
        CONVERT(varchar, t0.U_ORC_BE_ProdTime, 108) BETWEEN '07:00:00' AND '19:00:00' THEN '1'
    ELSE '2'
END  AS [Shift],
    --If time is between 7am and 7pm then Shift 1, else Shift 2

我想将它插入到我已经拥有的查询中，然后让它也按班次分组。

    SELECT 

T6.U_ORC_BE_StyleName AS [Style Name],
T0.[ItemCode],
STUFF(T0.[ItemCode], 1, 6, '') AS [Shortened ItemCode],
SUM(T0.[PlannedQty]) AS [Planned Qty],
SUM(T0.CmpltQty) AS [Actual Qty], 
SUM(T0.CmpltQty) - SUM(T0.PlannedQty) AS [Qty Difference],
SUM(T5.U_ORC_BE_HECTOLITER * T0.CmpltQty) AS [Total Hectoliters],

CASE
    WHEN [.......]
    THEN
    WHEN [.......]
    THEN
END AS [Line],

T0.U_ORC_BE_ProdDate AS [Date Produced]

    FROM [.......]

    GROUP BY T6.U_ORC_BE_StyleName, T0.[ItemCode], T5.Code , T0.U_ORC_BE_ProdDate

它不断引发错误，但我怀疑是由于尝试按别名分组。

＆＃34;每个GROUP BY表达式必须包含至少一个不是外部引用的列。＆＃34;

Answer 1

要按计算列（如case语句）进行分组，您还必须在该组中包含该语句。例如：

SELECT
  CASE ... END AS CaseStatement,
  [YourColumns]
FROM YourTable
GROUP BY CASE ... END

Answer 2

如错误所述，您无法按别名分组。在你的小组之后你必须复制整个案例。

 select 
 ....
 group by 
 T6.U_ORC_BE_StyleName, T0.[ItemCode], T5.Code , T0.U_ORC_BE_ProdDate,
 CASE
    WHEN [.......]
    THEN
    WHEN [.......]
    THEN
END

SQL GROUP BY别名

2 个答案: