Question

我希望在sql语句

中的case语句中进行比较

  SUM(CASE 
    WHEN bt.idBillingStatus = 2  and o.idOperator = 5 and cs.idClub=1
    THEN (price * revenueShareAmountLocal * o.mfactor *0.08)

，第二个是

SUM(CASE 
    WHEN bt.idBillingStatus = 2  and o.idOperator = 5 and cs.idClub=2
    THEN (price * revenueShareAmountLocal * o.mfactor *0.04)

我想进行if bt.idBillingStatus = 2 and o.idOperator = 5 and cs.idClub=2 > bt.idBillingStatus = 2 and o.idOperator = 5 and cs.idClub=1的比较然后(price * revenueShareAmountLocal * o.mfactor *0.01)

否则 if bt.idBillingStatus = 2 and o.idOperator = 5 and cs.idClub=2 < bt.idBillingStatus = 2 and o.idOperator = 5 and cs.idClub=1 然后(price * revenueShareAmountLocal * o.mfactor *0.08)

简单来说，我需要在案例陈述中进行比较，如果一个大于另一个，那么将其他总和加上其他东西。

Answer 1

由于您未包含数据或表格结构，我们在此处提供的帮助有限，但听起来您需要以下其中一项：

SUM(CASE 
      WHEN bt.idBillingStatus = 2  and o.idOperator = 5 and cs.idClub=1
        THEN (price * revenueShareAmountLocal * o.mfactor *0.08)

      WHEN bt.idBillingStatus = 2  and o.idOperator = 5 and cs.idClub=2
        THEN (price * revenueShareAmountLocal * o.mfactor *0.04)

      ELSE 0
    END
   )

或者更加冗长的选项，这表明您应该查看错误的查询设计：

CASE
  WHEN SUM(CASE 
             WHEN bt.idBillingStatus = 2  and o.idOperator = 5 and cs.idClub=2
               THEN (price * revenueShareAmountLocal * o.mfactor *0.04)

             ELSE 0
           END
          )
       > SUM(CASE 
             WHEN bt.idBillingStatus = 2  and o.idOperator = 5 and cs.idClub=1
               THEN (price * revenueShareAmountLocal * o.mfactor *0.08)

             ELSE 0
           END
          )
  THEN SUM(CASE 
             WHEN bt.idBillingStatus = 2  and o.idOperator = 5 and cs.idClub=2
               THEN (price * revenueShareAmountLocal * o.mfactor *0.04)

             ELSE 0
           END
          )

   ELSE SUM(CASE 
             WHEN bt.idBillingStatus = 2  and o.idOperator = 5 and cs.idClub=1
               THEN (price * revenueShareAmountLocal * o.mfactor *0.08)

             ELSE 0
           END
          )
END

Answer 2

您可以在子查询中计算2个总和然后比较外部查询中的结果。这可能是最易读的方法。

或者你可以在总和之间加上一个案例。

例如：

-- Using a single table variable for testing reasons
declare @Test table (groupid int, idBillingStatus int, idOperator int, idClub int, price float, revenueShareAmountLocal float, mfactor int);
insert into @Test (groupid, idBillingStatus, idOperator, idClub, price, revenueShareAmountLocal, mfactor) values
(1000,2,5,1,100,20,2),
(1000,2,5,2,100,20,2),
(2000,2,5,2,100,20,2),
(2000,2,5,1,200,20,2),
(2000,2,5,2,100,20,2);

select groupid, idBillingStatus, idOperator, result1, result2,
(case 
 when result2 > result1 then result2 * 0.01
 else result2 * 0.08
 end) as finalresult2
from (
    select groupid, idBillingStatus, idOperator, 
    SUM(
     case 
     when idClub = 1 and idBillingStatus = 2 and idOperator = 5
     then price * revenueShareAmountLocal * mfactor
     end) as result1,
    SUM(
     case
     when idClub = 2 and idBillingStatus = 2 and idOperator = 5
     then price * revenueShareAmountLocal * mfactor
     end) as result2
    from @Test
    group by groupid, idBillingStatus, idOperator
) q;

-- On in 1 select, comparing the sums
select groupid, idBillingStatus, idOperator, 
CASE 
WHEN SUM(case when idClub = 2 and idBillingStatus = 2 and idOperator = 5 then price * revenueShareAmountLocal * mfactor end) >
     SUM(case when idClub = 1 and idBillingStatus = 2 and idOperator = 5 then price * revenueShareAmountLocal * mfactor end)
THEN SUM(case when idClub = 2 and idBillingStatus = 2 and idOperator = 5 then price * revenueShareAmountLocal * mfactor end) * 0.01
ELSE SUM(case when idClub = 2 and idBillingStatus = 2 and idOperator = 5 then price * revenueShareAmountLocal * mfactor end) * 0.08
END as result          
from @Test
group by groupid, idBillingStatus, idOperator;

案例陈述中的比较条件

2 个答案: