代码之家 › 专栏 › 技术社区 › lakshminb7

在SQL存储过程中选择并合并表中的行

stored-procedures tsql sql-server sql

lakshminb7 · 技术社区 · 16 年前

具有架构为:id seqno name的临时表

ID-不唯一
seqno-int(可以是1、2或3)。将id+seqno排序为主键
姓名-任何文本

表中的示例数据如下

1 | 1 | RecordA  
2 | 1 | RecordB  
3 | 1 | RecordC  
1 | 2 | RecordD  
4 | 1 | RecordE  
5 | 1 | RecordF  
3 | 1 | RecordG

需要从此表中选择并输出

1 | RecordA/RecordD  
2 | RecordB  
3 | RecordC/RecordG  
4 | RecordE  
5 | RecordF

需要在没有光标的情况下执行此操作。

3 回复 | 直到 13 年前

Andomar 16 年前

如果SEQNO限于1、2、3:

select id, a.name + coalesce('/'+b.name, '') + coalesce('/'+c.name, '')
from myTable a
left outer join myTable b on a.id=b.id and b.seqno = 2
left outer join myTable c on a.id=c.id and c.seqno = 3
where a.seqno = 1;

如果seqno是开放式的,则可以部署递归CTE:

;with anchor as (
   select id, name, seqno
      from myTable
      where seqno=1)
, recursive as (
   select id, name, seqno
      from anchor
      union all
   select t.id, r.name + '/' + t.name, t.seqno
      from myTable t
      join recursive  r on t.id = r.id and r.seqno+1 = t.seqno)
select id, name from recursive;

Peter Radocchia 16 年前

如果你知道Seqno永远不会超过3个:

select Id, Names = stuff(
    max(case when SeqNo = 1 then '/'+Name else '' end)
  + max(case when SeqNo = 2 then '/'+Name else '' end)
  + max(case when SeqNo = 3 then '/'+Name else '' end)
  , 1, 1, '')
from table1 
group by Id

否则,类似这样的问题是任意数量项目的通用解决方案:

select Id, Names = stuff((
  select '/'+Name from table1 b
  where a.Id = b.Id order by SeqNo
  for xml path (''))
  , 1, 1, '')
from table1 a
group by Id

或者写一个clr uda。

编辑:相关表上的别名错误!

edit2:另一个版本,基于remus的递归示例。我想不出任何方法来选择只有每个ID的最后一个递归,不进行聚合或排序。有人知道吗?

;with
  myTable as (
    select * from (
      values 
        (1, 1, 'RecordA')  
      , (2, 1, 'RecordB')  
      , (3, 1, 'RecordC')  
      , (1, 2, 'RecordD')  
      , (4, 1, 'RecordE')  
      , (5, 1, 'RecordF')  
      , (3, 2, 'RecordG')
      ) a (Id, SeqNo, Name)
    )    
, anchor as (
    select id, name = convert(varchar(max),name), seqno
    from myTable where seqno=1
    )
, recursive as (
    select id, name, seqno
    from anchor
    union all
    select t.id, r.name + '/' + t.name, t.seqno
    from myTable t
    join recursive  r on t.id = r.id and r.seqno+1 = t.seqno
    )
select id, name = max(name) 
from recursive
group by id;
---- without aggregation, we get 7 rows:
--select id, name
--from recursive;

j0k gauthamp 13 年前

解决方案是好的。我有一个类似的问题,但这里我使用了两个不同的表。前任:

table1 

1 | 1 |
2 | 3 |
3 | 1 |
4 | 2 |
5 | 1 |
1 | 2 |
1 | 3 | 
4 | 1 |
5 | 2 |
2 | 2 |
4 | 3 |

table2 
1 | RecordA
2 | RecordB
3 | RecordC

我想从两个表中获取数据并以下面的格式显示。

1 | RecordA,RecordB,RecordC|
2 | RecordB,RecordC|
3 | RecordA |
4 | RecordA,RecordB,RecordC|
5 | RecordA,RecordB |