In SQL, how to select minimum value of a column and group by other columns?

Question

I have a lookup table below:

id  ref order
1   6   0
2   6   0
3   7   0
5   34  0
6   33  0
6   255 1
9   12  0
9   80  1
12  7   0
12  76  1
13  10  0
15  12  0
16  6   0
16  7   1
17  6   1
17  63  0
18  7   0
19  7   1
19  75  0
20  6   0
20  63  1

So in the lookup table (tab_lkp), it has column [id] (the IDs of entities), [ref] (the reference id that points to other entities in another table) and [order] (tells the order of reference, smaller order means higher priority).

My expectation is that, for each of the IDs, only one ref with the smallest order is selected. My code is (by following Phil's answer ):

select id
,      ref
,      min_order = min(order)
from [dbo].[tab_lkp]
group by id, ref
order by id, ref

But the code doesn't work for me, the results still contains multiple records for each of the IDs:

id  ref order
1   6   0
2   6   0
3   7   0
5   34  0
6   33  0
6   255 1
9   12  0
9   80  1
12  7   0
12  76  1
13  10  0
15  12  0
16  6   0
16  7   1
17  6   1
17  63  0
18  7   0
19  7   1
19  75  0
20  6   0
20  63  1

Could you please let me know what is wrong with my code? And how should I achieve my goal?

Answer 1

You would normally do this using row_number() :

select t.*
from (select t.*, row_number() over (partition by id order by ref) as seqnum
      from [dbo].[tab_lkp] t
     ) t
where seqnum = 1;

Answer 2

or by using a subquery that does exactly what you state that you want, "for each of the IDs, only one ref with the smallest order is selected"

Select * from tab_lkp t
Where order = 
    (Select Min(order) from tab_lkp 
     where Id = t.Id)

Answer 3

From an ANSI sql approach:

select x2.id, x2.ref, x2.order
from MyTable x2
inner join 
(
  select id, min(order) as min_order
  from MyTable
  group by id
) x1
on x1.id = x2.id
and x1.min_order = x2.order