我想写一个这样的查询:

SELECT o.OrderId, MAX(o.NegotiatedPrice, o.SuggestedPrice)
FROM Order o

但这不是MAX函数的工作原理,对吧?它是一个聚合函数,因此它需要一个参数,然后返回所有行的MAX。

有人知道我的方法吗?


当前回答

CREATE FUNCTION [dbo].[fnMax] (@p1 INT, @p2 INT)
RETURNS INT
AS BEGIN

    DECLARE @Result INT

    SET @p2 = COALESCE(@p2, @p1)

    SELECT
        @Result = (
                   SELECT
                    CASE WHEN @p1 > @p2 THEN @p1
                         ELSE @p2
                    END
                  )

    RETURN @Result

END

其他回答

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

扩展Xin的答案并假设比较值类型是INT,这种方法也适用:

SELECT IIF(ISNULL(@A, -2147483648) > ISNULL(@B, -2147483648), @A, @B)

这是一个完整的测试示例值:

DECLARE @A AS INT
DECLARE @B AS INT

SELECT  @A = 2, @B = 1
SELECT  IIF(ISNULL(@A, -2147483648) > ISNULL(@B, -2147483648), @A, @B)
-- 2

SELECT  @A = 2, @B = 3
SELECT  IIF(ISNULL(@A, -2147483648) > ISNULL(@B, -2147483648), @A, @B)
-- 3

SELECT  @A = 2, @B = NULL
SELECT  IIF(ISNULL(@A, -2147483648) > ISNULL(@B, -2147483648), @A, @B)
-- 2    

SELECT  @A = NULL, @B = 1
SELECT  IIF(ISNULL(@A, -2147483648) > ISNULL(@B, -2147483648), @A, @B)
-- 1

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;

我不这么想。我那天想要这个。我最接近的说法是:

SELECT
  o.OrderId,
  CASE WHEN o.NegotiatedPrice > o.SuggestedPrice THEN o.NegotiatedPrice 
     ELSE o.SuggestedPrice
  END
FROM Order o

对于SQL Server 2012:

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