1

我已经找到了这个有趣的线程

在 Microsoft SQL Server 2005 中模拟 group_concat MySQL 函数？

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

这给了我

1   5,2,5
2   1
3   5,2,2
4   5,3

但是我想从每次考试中删除重复项并对连接的结果进行排序以获得

1  2,5
2  1
3  2,5
4  3,5

谢谢。

4

2 回答 2

4

尝试在内部查询中使用 DISTINCT 并按方法而不是 id_exam 排序。

select 
id_exam, 
method = replace ((select distinct method AS [data()]
                   from methods
                   where id_exam = a.id_exam                      
                   order by method for xml path('')), ' ', ',')
from  methods a
where id_exam is not null
group by id_exam
于 2011-06-19T12:50:01.427 回答
1

将 a 添加group by method到内部查询并更改order by id_exam为order by method.

select 
id_exam, 
method = replace ((select method AS [data()]
                   from methods
                   where id_exam = a.id_exam
                   group by method                      
                   order by method for xml path('')), ' ', ',')
from  methods a
where id_exam is not null
group by id_exam

结果：

id_exam     method
----------- ---------
1           2,5
2           1
3           2,5
4           3,5
于 2011-06-19T13:00:08.223 回答