Order rows by values - sql

Question

I trying order table by rank, but rows which have position value - have to have position according to value in position field. It is possible do it without additional tables, views etc?
I have table like this:
rank | position | name
999 | 10 | txt1
200 | 4 | txt2
32 | 1 | txt3
1200 | 2 | txt4
123 | null | txt5
234 | null | txt6
567 | null | txt7
234 | null | txt8
432 | null | txt9
877 | null | txt10
Desired output have to look like this:
rank | position | name
32 | 1 | txt3
1200 | 2 | txt4
877 | null | txt10
200 | 4 | txt2
567 | null | txt7
432 | null | txt9
345 | null | txt8
234 | null | txt6
123 | null | txt5
999 | 10 | txt1

Answer 1

Here is an idea. Assign the proper ordering to each row. Then, if the position is available use that instead. When there are ties, put the position value first:
select t.*
from (select t.*, row_number() over (order by rank desc) as seqnum
from t
) t
order by (case when position is not null then position else seqnum end),
(case when position is not null then 1 else 2 end);
SQL Fiddle doesn't seem to be working these days, but this query demonstrates the results:
with t(rank, position, t) as (
select 999, 10, 'txt1' union all
select 200, 4, 'txt2' union all
select 32 , 1, 'txt3' union all
select 1200, 2, 'txt4' union all
select 123, null, 'txt5' union all
select 234, null, 'txt6' union all
select 567, null, 'txt7' union all
select 234, null, 'txt8' union all
select 432, null, 'txt9' union all
select 877, null , 'txt10'
)
select t.*
from (select t.*, row_number() over (order by rank desc) as seqnum
from t
) t
order by (case when position is not null then position else seqnum end),
(case when position is not null then 1 else 2 end);
EDIT;
When I wrote the above, I had a nagging suspicion of a problem. Here is a solution that should work. It is more complicated but it does produce the right numbers:
with t(rank, position, t) as (
select 999, 10, 'txt1' union all
select 200, 4, 'txt2' union all
select 32 , 1, 'txt3' union all
select 1200, 2, 'txt4' union all
select 123, null, 'txt5' union all
select 234, null, 'txt6' union all
select 567, null, 'txt7' union all
select 234, null, 'txt8' union all
select 432, null, 'txt9' union all
select 877, null , 'txt10'
)
select *
from (select t.*, g.*,
row_number() over (partition by t.position order by t.rank) gnum
from generate_series(1, 10) g(n) left join
t
on t.position = g.n
) tg left join
(select t.*,
row_number() over (partition by t.position order by t.rank) as tnum
from t
) t
on tg.gnum = t.tnum and t.position is null
order by n;
This is a weird sort of interleaving problem. The idea is to create slots (using generate series) for the positions. Then, assign the known positions to the slots. Finally, enumerate the remaining slots and assign the values there.
Note: I hard-coded 10, but it is easy enough to put in count(*) from the table there.

Answer 2

suppose you stored data in table1.
Then you should update column "position" as follows:
update a
set position = x.pos_null
from table1
a
inner join
(
select
a.name,
COUNT(a.rank) as pos_null
from
(
select
*
from table1
where position is null
)
a
left join
(
select
*
from table1
)
b
on a.rank <= b.rank
group by
a.name
)
x
on a.name = x.name
select * from table1 order by position
Bye,
Angelo.