SQL Server 2008中 - 透視

Question

create table [temp](
[id] [nvarchar](10) not null, 
[name] [nvarchar](50) not null, 
[info1] [nvarchar](50) not null, 
[info2] [nvarchar](50) not null, 
[info3] [nvarchar](50) not null); 

insert into temp(id,name,info1,info2,info3) values ('id1','name1','infoa','infob','infoc'); 
insert into temp(id,name,info1,info2,info3) values ('id1','name1','infox','infod','infoc'); 
insert into temp(id,name,info1,info2,info3) values ('id1','name1','infoz','infob','infoc'); 

表看起來遵循從臨時表

temp table 
id   name info1  info2  info3 
id1  name1 infoa  infob  infoc 
id1  name1 infox  infod  infoc 
id1  name1 infoy  infob  infoc 

多行將由ID，名稱和所有獨特的信息欄分組將被串聯預期輸出

id name info1    info2   info3 
id1 name1 infoa;infox;infoy infob;infod infoc 

Answer 1

select [id],[name], 
    stuff((select distinct ',' + CAST(t2.[info1] as varchar(10)) 
    from [temp] t2 where t1.id = t2.id and t1.name = t2.name 
    for xml path('')),1,1,'') info1, 
    stuff((select distinct ',' + CAST(t3.[info2] as varchar(10)) 
    from [temp] t3 where t1.id = t3.id and t1.name = t3.name 
    for xml path('')),1,1,'') info2, 
    stuff((select distinct ',' + CAST(t4.[info3] as varchar(10)) 
    from [temp] t4 where t1.id = t4.id and t1.name = t4.name 
    for xml path('')),1,1,'') info3 
from [temp] t1 
group by id,Name 

使用的功能

1. Stuff
2. For XML
3. Group by