DECLARE @TableName TABLE (Number INT, Date1 DATETIME, Date2 DATETIME, Date3 DATETIME, Cost MONEY)

INSERT INTO @TableName 
SELECT 1, '20000101', '20010101','20020101',100 UNION ALL
SELECT 2, '20000101', '19900101','19980101',99 

SELECT Number,
       Cost  ,
       (SELECT MAX([Date])
       FROM    (SELECT Date1 AS [Date]
               UNION ALL
               SELECT Date2
               UNION ALL
               SELECT Date3
               )
               D
       )
       [Most Recent Date]
FROM   @TableName

使用CROSS APPLY (2005+) ....

SELECT MostRecentDate 
FROM SourceTable
    CROSS APPLY (SELECT MAX(d) MostRecentDate FROM (VALUES (Date1), (Date2), (Date3)) AS a(d)) md

如果你正在使用MySQL或PostgreSQL或Oracle或BigQuery，你可以使用

SELECT GREATEST(col1, col2 ...) FROM table

我的解决方案也可以处理空值比较。它可以被简化为一个单一的查询，但为了解释，我使用CTE。这个想法是在第一步中减少从3个数字到2个数字的比较，然后在第二步中从2个数字到1个数字的比较。

with x1 as
(
  select 1 as N1, null as N2, 3 as N3
  union
  select 1 as N1, null as N2, null as N3
  union
  select null as N1, null as N2, null as N3
)
,x2 as
(
select 
N1,N2,N3,
IIF(Isnull(N1,0)>=Isnull(N2,0),N1,N2) as max1,
IIF(Isnull(N2,0)>=Isnull(N3,0),N2,N3) as max2
from x1
)
,x3 as
(
 select N1,N2,N3,max1,max2,
 IIF(IsNull(max1,0)>=IsNull(max2,0),max1,max2) as MaxNo
 from x2
)
select * from x3

输出:

如果您正在使用SQL Server 2005，您可以使用UNPIVOT特性。下面是一个完整的例子:

create table dates 
(
  number int,
  date1 datetime,
  date2 datetime,
  date3 datetime 
)

insert into dates values (1, '1/1/2008', '2/4/2008', '3/1/2008')
insert into dates values (1, '1/2/2008', '2/3/2008', '3/3/2008')
insert into dates values (1, '1/3/2008', '2/2/2008', '3/2/2008')
insert into dates values (1, '1/4/2008', '2/1/2008', '3/4/2008')

select max(dateMaxes)
from (
  select 
    (select max(date1) from dates) date1max, 
    (select max(date2) from dates) date2max,
    (select max(date3) from dates) date3max
) myTable
unpivot (dateMaxes For fieldName In (date1max, date2max, date3max)) as tblPivot

drop table dates

上表为员工工资表，列为salary1、salary2、salary3、salary4。下面的查询将返回四列中的最大值

select  
 (select Max(salval) from( values (max(salary1)),(max(salary2)),(max(salary3)),(max(Salary4)))alias(salval)) as largest_val
 from EmployeeSalary

运行上述查询将输出largest_val(10001)

上述查询逻辑如下:

select Max(salvalue) from(values (10001),(5098),(6070),(7500))alias(salvalue)

输出将是10001