对于上面关于大数字的答案，你可以在加减法之前做乘法。它有点笨重，但不需要石膏。(我不能说速度，但我认为它仍然非常快)

选择0.5*((@val1+@val2)+ ABS(@val1 - @val2)

更改

选择 @val1*0.5+@val2*0.5 + ABS（@val1*0.5 - @val2*0.5）

如果你想避免强制转换，至少有一个替代方案。

SELECT o.OrderID
CASE WHEN o.NegotiatedPrice > o.SuggestedPrice THEN
 o.NegotiatedPrice
ELSE
 o.SuggestedPrice
END AS Price

你可以这样做:

select case when o.NegotiatedPrice > o.SuggestedPrice 
then o.NegotiatedPrice
else o.SuggestedPrice
end

其实很简单:

CREATE FUNCTION InlineMax
(
    @p1 sql_variant,
    @p2 sql_variant
)  RETURNS sql_variant
AS
BEGIN
    RETURN CASE 
        WHEN @p1 IS NULL AND @p2 IS NOT NULL THEN @p2 
        WHEN @p2 IS NULL AND @p1 IS NOT NULL THEN @p1
        WHEN @p1 > @p2 THEN @p1
        ELSE @p2 END
END;

DECLARE @MAX INT
@MAX = (SELECT MAX(VALUE) 
               FROM (SELECT 1 AS VALUE UNION 
                     SELECT 2 AS VALUE) AS T1)

对于SQL Server 2012:

SELECT 
    o.OrderId, 
    IIF( o.NegotiatedPrice >= o.SuggestedPrice,
         o.NegotiatedPrice, 
         ISNULL(o.SuggestedPrice, o.NegiatedPrice) 
    )
FROM 
    Order o