SQL Server四舍五入问题，其中有5

Question

据我所知，当四舍五入为5时，四舍五入应按以下方式进行。

2.435 => 2.44 (Round Up, if rounding to digit(3) is odd number)
2.445 => 2.44 (Round Down, if rounding to digit(4) is even number)

如果我们做所有的总结，

2.435 + 2.445 =  4.88
2.44 + 2.44 = 4.88

我很确定在.Net中也可以进行这种四舍五入。

但是在SQL Server中，总会四舍五入，这在数学上是不正确的。

SELECT round(2.345, 2)  = 2.35
SELECT round(2.335, 2) => 2.34

这导致四舍五入后的总值相差1美分。

2.345 + 2.335 = 4.68
2.35 + 2.34 = 4.69 => which is not correct

我已经尝试过使用十进制和货币数据类型。

难道我做错了什么？有没有解决的办法？

Answer 1

如果您确实想在SQL Server中使用银行取整…

CREATE FUNCTION BankersRounding(@value decimal(36,11), @significantDigits INT)        
RETURNS MONEY        
AS        
BEGIN        
    -- if value = 12.345 and signficantDigits = 2...

    -- base = 1000        
    declare @base int = power(10, @significantDigits + 1)


    -- roundingValue = 12345        
    declare @roundingValue decimal(36,11) = floor(abs(@value) * @base)        
    -- roundingDigit = 5        
    declare @roundingDigit int = @roundingValue % 10

    -- significantValue = 1234        
    declare @significantValue decimal(36,11) = floor(@roundingValue / 10)        
    -- lastSignificantDigit = 4        
    declare @lastSignificantDigit int = @significantValue % 10


    -- awayFromZero = 12.35        
    declare @awayFromZero money = (@significantValue + 1) / (@base / 10)        
    -- towardsZero = 12.34        
    declare @towardsZero money = @significantValue / (@base / 10)

    -- negative values handled slightly different        
    if @value < 0        
    begin        
        -- awayFromZero = -12.35        
        set @awayFromZero = ((-1 * @significantValue) - 1) / (@base / 10)        
        -- towardsZero = -12.34        
        set @towardsZero = (-1 * @significantValue) / (@base / 10)        
    end

    -- default to towards zero (i.e. assume thousandths digit is 0-4)        
    declare @rv money = @towardsZero        
    if @roundingDigit > 5        
        set @rv = @awayFromZero  -- 5-9 goes away from 0        
    else if @roundingDigit = 5         
    begin        
        -- 5 goes to nearest even number (towards zero if even, away from zero if odd)        
        set @rv = case when @lastSignificantDigit % 2 = 0 then @towardsZero else @awayFromZero end        
    end

    return @rv

end