我已经发现了这个有趣的线程删除重复和排序结果模拟GROUP_CONCAT
Simulating group_concat MySQL function in Microsoft SQL Server 2005?
use test
go
create table methods (
id int identity,
id_exam int,
method int
)
go
insert into methods (id_exam,method) values (1,5)
insert into methods (id_exam,method) values (1,2)
insert into methods (id_exam,method) values (1,5)
insert into methods (id_exam,method) values (2,1)
insert into methods (id_exam,method) values (3,5)
insert into methods (id_exam,method) values (3,2)
insert into methods (id_exam,method) values (3,2)
insert into methods (id_exam,method) values (4,5)
insert into methods (id_exam,method) values (4,3)
select
id_exam,
method = replace ((select method AS [data()]
from methods
where id_exam = a.id_exam
order by id_exam for xml path('')), ' ', ',')
from methods a
where id_exam is not null
group by id_exam
,让我
1 5,2,5
2 1
3 5,2,2
4 5,3
但是我想从每个删除重复考试和排序连接结果以获得
1 2,5
2 1
3 2,5
4 3,5
谢谢。
非常感谢你。它效果很好。 :) –