select
   datediff(mi,b.start_time ,a.end_time)
from
   (select id=row_number()over(order by getdate()),* from tb)a,
   (select id=row_number()over(order by getdate()),* from tb)b
where
   a.id=b.id-1

------解决方案--------------------
SQL code
----------------------------------------------
-- Author : htl258(Tony)
-- Date   : 2010-05-10 11:37:00
-- Version: Microsoft SQL Server 2008 (RTM) - 10.0.1600.22 (Intel X86) 
--          Jul  9 2008 14:43:34 
--          Copyright (c) 1988-2008 Microsoft Corporation
--          Developer Edition on Windows NT 5.1 <X86> (Build 2600: Service Pack 3)
-- Blog   : http://blog.csdn.net/htl258
----------------------------------------------

--> 生成测试数据表: [tb]
IF OBJECT_ID('[tb]') IS NOT NULL
    DROP TABLE [tb]
GO
CREATE TABLE [tb] ([job] [nvarchar](20),[gx_id] [int],[start_time] [datetime],[end_time] [datetime])
INSERT INTO [tb]
SELECT '0001011-3-1','10','2010-5-10 10:00:00','2010-5-10 15:23:10' UNION ALL
SELECT '0002695-1-2','30','2010-5-10 15:30:00','2010-5-10 18:00:00' UNION ALL
SELECT '0003625-6-2','60','2010-5-11 07:20:05','2010-5-11 11:30:00' UNION ALL
SELECT '0009658-8-1','20','2010-5-12 09:20:30','2010-5-13 12:10:00'

--SELECT * FROM [tb]
--我想得到表A第一行的 "end_time "与第二行的 "start_time "相减的值，第二行的 "end_time "与第三行的 "start_time "相减的值，
--第三行的 "end_time "与第四行的 "start_time "相减的值，以此类推，请问怎样实现呢？
-->SQL查询如下:
;with t as
(
    select rn=row_number()over(ORDER by job),*
    from tb
)
select a.*,val=datediff(mi,a.end_time,b.bstart_time)
from t a
    outer apply(
        select bstart_time=start_time 
        from t 
        where a.rn=rn-1
        ) b
/*
rn                   job                  gx_id       start_time              end_time                val
-------------------- -------------------- ----------- ----------------------- ----------------------- -----------
1                    0001011-3-1          10          2010-05-10 10:00:0