没有内置函数返回两个（或更多）列的最小/最大值。您可以实现自己的标量函数来执行此操作。

在SQL Server 2005+中，您可以使用 UNPIVOT 将列转换成行，然后使用MIN函数：

  CREATE TABLE [X] 
（
 [ID] INT，
 [Date1] DATETIME，
 [Date2] DATETIME，
 [Date3] DATETIME 
）

 INSERT [X] 
 VALUES（0，'09 / 29/2011'，'09 / 20/2011'，'09 / 01/2011'），
（1，'01 / 01/2011'，'01/05/2011'，'03 / 03/2010'）


 SELECT [ID]，MIN（[Date] ）AS [MinDate] 
 FROM [X] 
 UNPIVOT（
 [日期] FOR d IN 
（[Date1] 
，[Date2] 
， [Date3]）
）unpvt 
 GROUP BY [ID] 
 

I'm using SQL Server 2008;

Suppose I have a table 'X' with columns 'Date1', 'Date2', 'Dateblah', all of type DateTime.

I want to select the min value between the three columns, for example (simplified, with date mm/dd/yyyy)

ID       Date1          Date2           Dateblah
0     09/29/2011      09/20/2011       09/01/2011 
1     01/01/2011      01/05/2011       03/03/2010

ID    MinDate
0    09/01/2011
1    03/03/2010

Is there a bread and butter command to do that ?

Thanks in advance.

EDIT: I've seen this question What's the best way to select the minimum value from multiple columns? but unfortunately it won't suit me as I'm being obligated to do it against normalization because I'm making tfs work item reports, and the 'brute-force' case thing will end up being a pain if I have 6 ou 7 columns.

There is no built in function to return the min/max of two (or more) columns. You could implement your own scalar function to do this.

In SQL Server 2005+ you could use UNPIVOT to turn the columns into rows and then use the MIN function:

CREATE TABLE [X]
(
    [ID] INT,
    [Date1] DATETIME,
    [Date2] DATETIME,
    [Date3] DATETIME
)

INSERT  [X]
VALUES  (0, '09/29/2011', '09/20/2011', '09/01/2011'),
        (1, '01/01/2011', '01/05/2011', '03/03/2010')


SELECT [ID], MIN([Date]) AS [MinDate]
FROM [X]
UNPIVOT (
    [Date] FOR d IN
        ([Date1]
        ,[Date2]
        ,[Date3])
) unpvt
GROUP BY [ID]