SELECT o.OrderId,   
--MAX(o.NegotiatedPrice, o.SuggestedPrice)  
(SELECT MAX(v) FROM (VALUES (o.NegotiatedPrice), (o.SuggestedPrice)) AS value(v)) as ChoosenPrice  
FROM Order o

可以在一行中完成:

-- the following expression calculates ==> max(@val1, @val2)
SELECT 0.5 * ((@val1 + @val2) + ABS(@val1 - @val2)) 

编辑:如果处理的是非常大的数字，则必须将值变量转换为bigint，以避免整数溢出。

对于SQL Server 2012:

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

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。

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)

我可能不会这样做，因为它比前面提到的CASE结构效率更低——除非您为两个查询都有覆盖索引。不管怎样，对于类似的问题，这都是一个有用的技巧:

SELECT OrderId, MAX(Price) as Price FROM (
   SELECT o.OrderId, o.NegotiatedPrice as Price FROM Order o
   UNION ALL
   SELECT o.OrderId, o.SuggestedPrice as Price FROM Order o
) as A
GROUP BY OrderId