在SQL Server 2012或更高版本中，您可以使用IIF和ISNULL(或COALESCE)的组合来获得最多2个值。 即使其中一个是NULL。

IIF(col1 >= col2, col1, ISNULL(col2, col1)) 

或者当两者都为NULL时，希望它返回0

IIF(col1 >= col2, col1, COALESCE(col2, col1, 0)) 

示例代码片段:

-- use table variable for testing purposes
declare @Order table 
(
  OrderId int primary key identity(1,1),
  NegotiatedPrice decimal(10,2),
  SuggestedPrice decimal(10,2)
);

-- Sample data
insert into @Order (NegotiatedPrice, SuggestedPrice) values
(0, 1),
(2, 1),
(3, null),
(null, 4);

-- Query
SELECT 
     o.OrderId, o.NegotiatedPrice, o.SuggestedPrice, 
     IIF(o.NegotiatedPrice >= o.SuggestedPrice, o.NegotiatedPrice, ISNULL(o.SuggestedPrice, o.NegotiatedPrice)) AS MaxPrice
FROM @Order o

结果:

OrderId NegotiatedPrice SuggestedPrice  MaxPrice
1       0,00            1,00            1,00
2       2,00            1,00            2,00
3       3,00            NULL            3,00
4       NULL            4,00            4,00

但如果需要多列中的最大值呢? 然后我建议在值的聚合上使用CROSS APPLY。

例子:

SELECT t.*
, ca.[Maximum]
, ca.[Minimum], ca.[Total], ca.[Average]
FROM SomeTable t
CROSS APPLY (
   SELECT 
    MAX(v.col) AS [Maximum], 
    MIN(v.col) AS [Minimum], 
    SUM(v.col) AS [Total], 
    AVG(v.col) AS [Average]
   FROM (VALUES (t.Col1), (t.Col2), (t.Col3), (t.Col4)) v(col)
) ca

这有一个额外的好处，它可以同时计算其他的东西。

哎呀，我刚刚发布了一个关于这个问题的恶搞帖…

答案是，没有像Oracle's Greatest这样的内置函数，但是您可以通过UDF为两个列实现类似的结果，注意，sql_variant的使用在这里非常重要。

create table #t (a int, b int) 

insert #t
select 1,2 union all 
select 3,4 union all
select 5,2

-- option 1 - A case statement
select case when a > b then a else b end
from #t

-- option 2 - A union statement 
select a from #t where a >= b 
union all 
select b from #t where b > a 

-- option 3 - A udf
create function dbo.GREATEST
( 
    @a as sql_variant,
    @b as sql_variant
)
returns sql_variant
begin   
    declare @max sql_variant 
    if @a is null or @b is null return null
    if @b > @a return @b  
    return @a 
end


select dbo.GREATEST(a,b)
from #t

克里斯汀

下面是我的回答:

create table #t (id int IDENTITY(1,1), a int, b int)
insert #t
select 1,2 union all
select 3,4 union all
select 5,2

select id, max(val)
from #t
    unpivot (val for col in (a, b)) as unpvt
group by id

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)

select OrderId, (
    select max([Price]) from (
        select NegotiatedPrice [Price]
        union all
        select SuggestedPrice
    ) p
) from [Order]

子查询可以从Outer查询访问列，因此可以使用这种方法跨列使用MAX等聚合。(不过，当涉及到大量列时可能更有用)

;WITH [Order] AS
(
SELECT 1 AS OrderId, 100 AS NegotiatedPrice, 110 AS SuggestedPrice UNION ALL
SELECT 2 AS OrderId, 1000 AS NegotiatedPrice, 50 AS SuggestedPrice
)
SELECT
       o.OrderId, 
       (SELECT MAX(price)FROM 
           (SELECT o.NegotiatedPrice AS price 
            UNION ALL SELECT o.SuggestedPrice) d) 
        AS MaxPrice 
FROM  [Order]  o

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