有谁可以实现这个查询功能

Question

table1
ndb004         ndb005 ndb006 NDB007
20130302 660221 1
20130303 660102 -1  103
20130303 660204 1
20130303 660206 1
20130303 660204 -1
20130303 660221 1
20130303 660204 1
20130303 100101 -1  201在数据库里直接得到这个表talbe2
ndb004               ndb005                              ndb006
20130302 660221 -1
20130303 660221,660204,660206,660204,660221,660204 -1



sql语句

Answer 1

----------------------------------------------------------------
-- Author  :DBA_Huangzj(發糞塗牆)
-- Date    :2013-05-28 11:42:30
-- Version:
--      Microsoft SQL Server 2008 R2 (SP1) - 10.50.2500.0 (X64) 
-- Jun 17 2011 00:54:03 
-- Copyright (c) Microsoft Corporation
-- Enterprise Edition (64-bit) on Windows NT 6.1 <X64> (Build 7601: Service Pack 1, v.721)
--
----------------------------------------------------------------
--> 测试数据：[table1]
if object_id('[table1]') is not null drop table [table1]
go 
create table [table1]([ndb004] datetime,[ndb005] int,[ndb006] int,[NDB007] int)
insert [table1]
select '20130302',660221,1,null union all
select '20130303',660102,-1,103 union all
select '20130303',660204,1,null union all
select '20130303',660206,1,null union all
select '20130303',660204,-1,null union all
select '20130303',660221,1,null union all
select '20130303',660204,1,null union all
select '20130303',100101,-1,201
--------------开始查询--------------------------SELECT  a.[ndb004] ,
       
        STUFF(( SELECT  ',' + CONVERT(VARCHAR(128),[ndb005])
                FROM    [table1] b
                WHERE   b.[ndb004] = a.[ndb004]
                        AND b.[ndb006] = a.[ndb006]
              FOR
                XML PATH('')
              ), 1, 1, '') '[ndb005]',  a.[ndb006]
FROM    [table1] a
GROUP BY a.[ndb004] ,
        a.[ndb006] ----------------结果----------------------------
/* 
ndb004                  [ndb005]                                                                                                                                                                                                                                                         ndb006
----------------------- ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- -----------
2013-03-02 00:00:00.000 660221                                                                                                                                                                                                                                                           1
2013-03-03 00:00:00.000 660102,660204,100101                                                                                                                                                                                                                                             -1
2013-03-03 00:00:00.000 660204,660206,660221,660204    
*/

Answer 2

SELECT  a.[ndb004],a.[ndb006],               
stuff(( SELECT  ',' + CONVERT(VARCHAR(128),[ndb005]) FROM [table1] b WHERE b.[ndb004] = a.[ndb004] AND b.[ndb006] = a.[ndb006] FOR XML PATH('')), 1, 1, '') '[ndb005]' 
FROM [table1] a 
GROUP BY a.[ndb004],a.[ndb006]
很常见

Answer 3

借用下版主大佬的数据if object_id('[table1]') is not null drop table [table1]
go 
create table [table1]([ndb004] datetime,[ndb005] int,[ndb006] int,[NDB007] int)
insert [table1]
select '20130302',660221,1,null union all
select '20130303',660102,-1,103 union all
select '20130303',660204,1,null union all
select '20130303',660206,1,null union all
select '20130303',660204,-1,null union all
select '20130303',660221,1,null union all
select '20130303',660204,1,null union all
select '20130303',100101,-1,201
--------------开始查询--------------------------
select ndb004,ndb005=
stuff((select ','+cast(ndb005 as varchar) from table1 where ndb004=t.ndb004 for XML path('')),1,1,''),
MIN(ndb006) ndb006
from table1 t group by ndb004