对于SQL Server 2012:

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

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;

是的，有。

T-SQL (SQL Server 2022 (16.x))现在支持最大/最小函数:

MAX/MIN作为非聚合函数 现在Azure SQL数据库和SQL托管实例都支持这个功能。它将滚动到下一个版本的SQL Server。

逻辑函数- GREATEST (Transact-SQL) 此函数返回一个或多个表达式列表中的最大值。 最伟大的表达，…n)

在这种情况下:

SELECT o.OrderId, GREATEST(o.NegotiatedPrice, o.SuggestedPrice)
FROM [Order] o;

db < > fiddle演示

如果你使用的是SQL Server 2008(或更高版本)，那么这是更好的解决方案:

SELECT o.OrderId,
       (SELECT MAX(Price)
        FROM (VALUES (o.NegotiatedPrice),(o.SuggestedPrice)) AS AllPrices(Price))
FROM Order o

所有的信用和投票都应该去Sven对一个相关问题的答案，“多列的SQL MAX ?” 我说这是“最佳答案”，因为:

It doesn't require complicating your code with UNION's, PIVOT's, UNPIVOT's, UDF's, and crazy-long CASE statments. It isn't plagued with the problem of handling nulls, it handles them just fine. It's easy to swap out the "MAX" with "MIN", "AVG", or "SUM". You can use any aggregate function to find the aggregate over many different columns. You're not limited to the names I used (i.e. "AllPrices" and "Price"). You can pick your own names to make it easier to read and understand for the next guy. You can find multiple aggregates using SQL Server 2008's derived_tables like so: SELECT MAX(a), MAX(b) FROM (VALUES (1, 2), (3, 4), (5, 6), (7, 8), (9, 10) ) AS MyTable(a, b)

试试这个。它可以处理2个以上的值

SELECT Max(v) FROM (VALUES (1), (2), (3)) AS value(v)

在MemSQL中执行以下操作:

-- DROP FUNCTION IF EXISTS InlineMax;
DELIMITER //
CREATE FUNCTION InlineMax(val1 INT, val2 INT) RETURNS INT AS
DECLARE
  val3 INT = 0;
BEGIN
 IF val1 > val2 THEN
   RETURN val1;
 ELSE
   RETURN val2;
 END IF; 
END //
DELIMITER ;

SELECT InlineMax(1,2) as test;