--> 测试数据：[ta]
if object_id('[ta]') is not null drop table [ta]
go 
create table [ta]([id] int,[name] varchar(4))
insert [ta]
select 1,'张三' union all
select 2,'李四' union all
select 3,'王五' union all
select 4,'赵六'

--> 测试数据：[tb]
if object_id('[tb]') is not null drop table [tb]
go 
create table [tb]([id] int,[name] varchar(5))
insert [tb]
select 1,'1,2,4' union all
select 2,'2,3'
--------------开始查询--------------------------
go 
create function f_hb(@id varchar(10))
returns varchar(1000)
as
begin
  declare @str varchar(1000)
  set @str=''
  select @str=@str+','+[name] from [ta] where charindex(','+cast(id as varchar)+',',','+@id+',')>0
  return stuff(@str,1,1,'')
end
go 

select id,name=dbo.f_hb([name]) from [tb]

--drop function f_hb
--drop table ta,tb
/*

------解决方案--------------------
SQL code
--> 测试数据：[ta]
if object_id('[ta]') is not null drop table [ta]
go 
create table [ta]([id] int,[name] varchar(4))
insert [ta]
select 1,'张三' union all
select 2,'李四' union all
select 3,'王五' union all
select 4,'赵六'

--> 测试数据：[tb]
if object_id('[tb]') is not null drop table [tb]
go 
create table [tb]([id] int,[name] nvarchar(5),tab1 varchar(20))
insert [tb]
select 1,N'作者','1,2,4' union all
select 2,N'文章','2,3'
select b.ID,
       b.name,
       stuff((select ','+[name] 
       from ta a 
       where CHARINDEX(','+ltrim(a.id)+',',','+b.tab1+',')>0
       for XML path('')),1,1,'')
from tb b

------解决方案--------------------
探讨


你写的很神奇。
我都没这么写过。
------------------------------
前提是tab1的数据有可能增长，tab2的数据也可能添加新的 东西。


------解决方案--------------------
create table t1
(
	id int,
	name varchar(10)
)
insert into t1
select 1, '张三' union all
select 2, '李四' union all
select 3, '王五' union all
select 4, '赵六'
create table t2
(
	id int,
	name varchar(10),
	tab1 varchar(10)
)
insert into t2
select 1, '作者', '1,2,4' union all
select 2, '文章', '2,3'
select * from t1
select * from t2

select *,STUFF((select ','+name from t1 where CHARINDEX(ltrim(t1.id),ltrim(t2.tab1))>0 for xml path('')),1,1,'') as name  
from t2


----------------------------
id	name	tab1	name
1	作者	1,2,4	张三,李四,赵六
2	文章	2,3	李四,王五

------解决方案--------------------
我想了个简单解法：
select a.*,(
	select name+','  
	from tab1  
	where charindex( cast(id as varchar),a.tab1)>0
   for xml path('')
) c  
from tab2 a

------解决方案--------------------
是个美女啊！

一般会了这个就牛b了
我也在学习中，多多交流！