Question

以下脚本由成员（OMG Ponies）为the question i posted last nigh t撰写。脚本中有语法错误我无法理解。有人可以试一试吗？它使用MS SQL Server 2000/2005/2008

select 
'Steve'     as name, 
'4/20/1960' as DOB,
'12456'     as agentID, 
'Smith'     as agentName
into #TABLE1

insert into #TABLE1
select 
'Steve'        as name, 
'4/20/1960'    as DOB,
'12456'        as agentID, 
'John'         as agentName

insert into #TABLE1
select 
'Steve'        as name, 
'4/20/1960'    as DOB,
'12456'        as agentID, 
'Lary'         as agentName

select * from #TABLE1

+---------+-----------+----------+------------------+
|  Name   |    DOB      | AgentID |  AgentName      | 
+---------+-----------+----------+------------------+
| Steve    | 4/20/1960 | 12456     | John           |   
+---------+-----------+----------+------------------+
| Steve    | 4/20/1960 | 12456     | Lary           | 
+---------+-----------+----------+------------------+
| Steve    | 4/20/1960 | 12456     | Smith          | 
+---------+-----------+----------+------------------+


+---------+-----------+----------+----------------------+
|  Name   |    DOB      | AgentID |    AgentName        | 
+---------+-----------+----------+----------------------+
| Steve    | 4/20/1960 | 4444     | John,Larry, Smith   | 
+---------+-----------+----------+----------------------+

SELECT DISTINCT 
       t.name, 
       t.dob, 
       t.agentid, 
       STUFF(ISNULL(SELECT ', ' + x.agentname 
                      FROM TABLE1 x 
                     WHERE x.agentid = t.agentid 
                  GROUP BY x.agentname 
                   FOR XML PATH ('')), ''), 1, 2, '') 
  FROM TABLE1 t

[错误]脚本行：1-10 -------------------------    关键字'SELECT'附近的语法不正确。    消息：156，等级：15，状态：1，程序：，行：5

[错误]脚本行：1-10 -------------------------    '）'附近的语法不正确。    消息：102，等级：15，状态：1，程序：，行：9

Answer 1

这是作品！

select 'Steve' as name, '4/20/1960' as DOB, '12456' as agentID, 'Smith' as agentName into #TABLE1

insert into #TABLE1 select 'Steve' as name, '4/20/1960' as DOB, '12456' as agentID, 'John' as agentName

insert into #TABLE1 select 'Steve' as name, '4/20/1960' as DOB, '12456' as agentID, 'Lary' as agentName

select * from #TABLE1

SELECT DISTINCT t.name, t.dob, t.agentid, STUFF(ISNULL((SELECT ', ' + x.agentname FROM #TABLE1 x WHERE x.agentid = t.agentid GROUP BY x.agentname FOR XML PATH ('')), ''), 1, 2, '') FROM #TABLE1 t
DROP TABLE #Table1

--+---------+-----------+----------+------------------+ | Name | DOB | AgentID | AgentName | +---------+-----------+----------+------------------+ | Steve | 4/20/1960 | 12456 | John |
--+---------+-----------+----------+------------------+ | Steve | 4/20/1960 | 12456 | Lary | +---------+-----------+----------+------------------+ | Steve | 4/20/1960 | 12456 | Smith | +---------+-----------+----------+------------------+

--+---------+-----------+----------+----------------------+ | Name | DOB | AgentID | AgentName | +---------+-----------+----------+----------------------+ | Steve | 4/20/1960 | 4444 | John,Larry, Smith | +---------+-----------+----------+----------------------+

将两个表合并到一个表中

1 个答案: