总和之和

Question

我希望使用db2获得一笔总和。我明显得到-204错误，因为我正在尝试使用别名来计算聚合。我不知道如何处理这个问题并希望得到一些帮助。

SELECT 
    DA.AMOUNT,
    cast(sum(cast(FAT.TRANSACTION_AMOUNT as real)) as integer) as time_used,
    (DA.AMOUNT - sum(FAT.TRANSACTION_AMOUNT)) as time_to_use,
    concat(((sum(FAT.TRANSACTION_AMOUNT)/DA.AMOUNT)*100),'%') AS Percentage_used,
    **sum(time_used)/SUM(DA.AMOUNT) AS Total_Percentage_Used**    

    FROM FACT_Table as FAT
        RIGHT DIM_AWARD DA ON FAT.DIM_ID = DA.DIM_ID
        group by DA.AMOUNT

Answer 1

一种方法是在表达式中替换金额：

SELECT DA.AMOUNT,
       cast(sum(cast(FAT.TRANSACTION_AMOUNT as real)) as integer) as time_used,
       (DA.AMOUNT - sum(FAT.TRANSACTION_AMOUNT)) as time_to_use,
       concat(((sum(FAT.TRANSACTION_AMOUNT)/DA.AMOUNT)*100),'%') AS Percentage_used,
       cast(sum(cast(FAT.TRANSACTION_AMOUNT as real)) as integer)/DA.AMOUNT AS Total_Percentage_Used  
FROM FACT_Table as FAT
     RIGHT DIM_AWARD DA ON FAT.DIM_ID = DA.DIM_ID
group by DA.AMOUNT;

由于DA.AMOUNT中有group by，因此您无需将其包含在sum()中。

如果您担心sum()为NULL，请使用coalesce()：     coalesce（cast（sum（cast（FAT.TRANSACTION_AMOUNT as real））为整数），0.0）/ DA.AMOUNT AS Total_Percentage_Used