0

我的表看起来与此类似：

LocalAmount Firm_id Date    StoreCard_Id
1000        5532    46,545  14124
2000        5325    46,545  13124
30000       25235   46,545  14141
1500        2525    46,545  14214
2134        25235   46,545  14241
144150      2525    46,545  41441

我可以使用以下代码在一年内对 localAmount 求和：

select sum(II2.LocalTAmountWithoutVAT) as "Celkem"
from IssuedInvoices2 II2
join IssuedInvoices II on II.id = II2.parent_id
join firms F on F.id = II.firm_id
left join StoreCards SC on II2.StoreCard_ID = SC.ID
left join ABI_StoreCardsWithMenu(sc.ID) SCWM on SCWM.OUT_StoreCard_ID = SC.id
where
     ib_decodedate_year(II.docdate$date) = EXTRACT(YEAR FROM CURRENT_DATE)
    and F.id = '@{_firmID}'
group by 
    out_menu1Text
order by
    out_menu1Text

但我需要在两年内做同样的事情并减去这两个 SUMS。它必须在一个 SQL 代码中。

抱歉对我的问题的解释不好，但我不知道如何更好地解释它。谢谢大家的回复。

4

1 回答 1

3

您可以使用条件聚合来做到这一点：

select sum(case when ib_decodedate_year(II.docdate$date) = EXTRACT(YEAR FROM CURRENT_DATE) 
                then II2.LocalTAmountWithoutVAT
                else 0
           end) as ThisYear,
       sum(case when ib_decodedate_year(II.docdate$date) = EXTRACT(YEAR FROM CURRENT_DATE) - 1
                then II2.LocalTAmountWithoutVAT
                else 0
           end) as LastYear,
       (sum(case when ib_decodedate_year(II.docdate$date) = EXTRACT(YEAR FROM CURRENT_DATE) 
                 then II2.LocalTAmountWithoutVAT
                 else 0
            end) -
        sum(case when ib_decodedate_year(II.docdate$date) = EXTRACT(YEAR FROM CURRENT_DATE) - 1
                then II2.LocalTAmountWithoutVAT
                else 0
           end)
       ) as Diff       
from IssuedInvoices2 II2 join
     IssuedInvoices II
     on II.id = II2.parent_id join
     firms F
     on F.id = II.firm_id left join
     StoreCards SC on II2.StoreCard_ID = SC.ID left join
     ABI_StoreCardsWithMenu(sc.ID) SCWM
     on SCWM.OUT_StoreCard_ID = SC.id
where F.id = '@{_firmID}'
group by  out_menu1Text
order by out_menu1Text;
于 2013-06-06T18:01:08.723 回答