所以我有下表:

ID |产品图片 300 |/300-01.jpg 300 |/300-02.jpg 301 |/301.jpg 302 |/302.jpg

每个 ID 可以有无限数量的图像.我需要将所有图像引用连接到一列中，并且无法生成以下输出:

There could be an unlimited number of images per ID. I need to concatenate all the image references into one column, and I am having trouble generating the following output:

ID |产品图片 300 |/300-01.jpg；/300-02.jpg; 301 |/301.jpg;

推荐答案

-- cte with test data
;with T (ID, Product_Image) as
(
select 300, '/300-01.jpg' union all
select 300, '/300-02.jpg' union all
select 301, '/301.jpg' union all
select 302, '/302.jpg'
)

select
  T.ID,
  (select T2.Product_Image+'; '
   from T as T2 
   where T.ID = T2.ID
   for xml path(''), type).value('.[1]', 'nvarchar(max)') as Product_Images
from T
group by T.ID

结果:

ID   Product_Images
---- -------------------------
300  /300-01.jpg; /300-02.jpg; 
301  /301.jpg; 
302  /302.jpg;