Select and merge rows in a table in SQL Stored procedure

Question

Have a temp table with schema: ID | SeqNo | Name

ID - Not unique
SeqNo - Int (can be 1,2 or 3). Sort of ID+SeqNo as Primary key
Name - Any text

And sam

Answer 1

If you know SeqNo will never be more than 3:

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

Otherwise, something like this is the generic solution to an arbitrary number of items:

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

Or write a CLR UDA.

Edit: had the wrong alias on the correlated table!

Edit2: another version, based on Remus's recursion example. I couldn't think of any way to select only the last recursion per Id, without aggregation or sorting. Anybody know?

;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;