求合并行的sql

[xielk]

例:table1
id num
01 01,02,03,04
02 05,06,07,08

输出为 01,02,03,04,05,06,07,08

[pyguang]

select ,(select num from table1 b where b.id=1)as num1,(select b.num from table1 b where b.id=2) as num2 from table1
未测试过不知行不行

[wwwwb]

selct a.*,b.* from tt a left join tt b on cast(a.id as decimal(5,0))+1=cast(b.id as decimal(5,0))

[$扫地僧$]

create table A
(
id varchar(10),
num varchar(100)
)
insert A
select '01','01,02,03,04' union
select '02','05,06,07,08'

DECLARE @T_SQL varchar(8000)
set @T_SQL=''
select @T_SQL=@T_SQL + '(select num from A where id=''' + id + ''')' + ' + '','' + ' from A
set @T_SQL='select ' + left(@T_SQL,len(@T_SQL)-len(' + '','' + '))
exec (@T_SQL)

[funsuzhou]

declare @Sum_num varchar(8000)
set @Sum_num=''
select @Sum_num=@Sum_num+num from table1 where ......
select @Sum_num

[funsuzhou]

declare @Sum_num varchar(8000)
set @Sum_num=''
select @Sum_num=@Sum_num+num+',' from table1 where ......
set @Sum_num=left(@Sum_num,len(@Sum_num)-1)
select @Sum_num

[mislrb]

declare @t table(id varchar(2), num varchar(50))
insert @t
select '01','01,02,03,04' union all
select '02', '05,06,07,08'
select newnum=(select num from @t where id='01')+','+(select num from @t where id='02')

/*
newnum
-----------------------
01,02,03,04,05,06,07,08
*/

[funsuzhou]

To mislrb(aben)：
你那是啥玩意儿？不具有通用性啊^_^
难道有100行你就有100个()加起来？

[cen123]

CREATE TABLE table1
(
id varchar(10),
num varchar(100)
)

INSERT table1 SELECT '01','01,02,03,04'
UNION SELECT '02','05,06,07,08,09'

DECLARE @num_Sum VARCHAR(8000)
set @num_Sum=''
SELECT @num_Sum=@num_Sum + a.num+',' FROM table1 a

SELECT left(@num_Sum,len(@num_Sum)-1)

DROP TABLE table1