Question

这个函数只是返回＆＃34; 1＆＃34;结果当我尝试用select选择结果时无论输入数字是什么。

Create Function dbo.LotValue (@IdAssetGroup as bit, @TdValueBase as real, @TdPrice as real, @IdRiskfactor as bit)
RETURNS real
as begin

DECLARE @Output AS real

    set @Output = 
        (case 
            when @IdAssetGroup in (1, 8) then 1
            when @IdAssetGroup in (2,3,4,6,7) then (@TdValueBase / @TdPrice)
            when @IdAssetGroup in (5) then
                    Case @IdRiskfactor
                        when 8 then 1 --BLFT
                        else 0.01
                    end
            when @IdAssetGroup in (9) then
                    case 
                        when @IdRiskfactor in (1,42,84,113) then @TdValueBase --AUD, EUR,NZD,GBP
                        when @IdRiskfactor in (25,62) then (@TdValueBase / @TdPrice) --NDF: BRL, KRW
                        else (@TdValueBase / @TdPrice) *-1 --others FX
                    end
        end)
    return @Output;
end

它只是返回＆＃34; 1＆＃34;：

  select dbo.lotvalue(5,20,5,42)

Answer 1

您每次都会遇到第一个案例（when @IdAssetGroup in (1, 8) then 1），因为@IdAssetGroup是bit，只能是1或0。

你也和@IdRiskFactor做同样的事情，我想这需要改变。我假设你希望这些是INT s。

CREATE FUNCTION dbo.LotValue 
    (
      @IdAssetGroup as INT, 
      @TdValueBase as real, 
      @TdPrice as real, 
      @IdRiskfactor as INT
    )
RETURNS real

Answer 2

这：@IdAssetGroup as bit

您可能需要int

SQL函数只返回“1”

2 个答案: