Intelligent Advisor

How about something like this?

  1  with t as (
  2     select rownum as rn1
  3     , row_number() over (order by rownum desc) rn2
  4     , trunc(count(*) over () / 2) as mid
  5     from all_objects where rownum <= 6
  6  )
  7  select rn1 as leftside
  8  , rn2 as rightside
  9  from t
 10  where rn1 <= mid
 11* order by rn1
SQL> /

  LEFTSIDE  RIGHTSIDE
---------- ----------
         1          6
         2          5
         3          4

cheers,
Anthony

with t as ( select level*101 as x from dual connect by level <= 7),
    tt as ( select x, row_number() over (order by x) - 1 as rn, ceil(count(*) over () / 2) as win from t)
select min(decode(trunc(rn/win), 0, x)) as x1,
       min(decode(trunc(rn/win), 1, x)) as x2
  from tt
  group by mod(rn, win)
  order by mod(rn, win)
;

           X1            X2
------------- -------------
          101           505
          202           606
          303           707
          404

It seems to me as if you try to force a client issue (layout thingie) onto the database server.

The displaying of select results should be handled by the client. Most client tools have enough capabilities to do such a thing. It is generally a bad idea to do the whole layouting in the select already.

Split the result set and print the second part in reversed order.

btw: Same idea goes for stuff like pivot querys.

Currently your MOD(t1.rn, ...) = MOD(t2.rn,...)

is what is giving you

1 4
2 5
3 6

because that's what you asked me for when I supplied this query in your previous thread.

Now MOD is the wrong approach. You want

1 COUNT
2 COUNT-1
3 COUNT-2

So replace the MOD comparison with

AND t2.rn(+) = (SELECT count(1) from member) - ( t1.rn - 1 )
ORDER BY t1.rn

The outer join is so that if there are an odd number of rows, t1 value will be present and t2 value will be null for the last row.

Message was edited by:
Dave Hemming

Using aketi's test table query would be

with t as
(SELECT val as nom, row_number() over (order by val ASC) as rn from testT)
select t1.nom as leftside, decode(t2.nom,t1.nom,null,t2.nom) as rightside
from t t1, t t2
where rownum < (SELECT COUNT(val) / 2 + 1 AS table_length FROM testt) 
AND t2.rn(+) = (SELECT count(1) from testt) - ( t1.rn - 1 )

This is an interesting question.

create table testT(Val) as
select RowNum*10 from all_catalog
 where RowNum <=7;

select Left,Right
from (select Rn,Cnt,Val as Left,
      case when Rn <= Cnt/2
           then Lead(Val,abs(Cnt-2*Rn+1)) over(order by Val) end as Right
      from (select Val,
            Row_Number() over(order by Val) as Rn,
            count(*) over() as Cnt
            from testT))
where Rn <= ceil(Cnt/2);

LEFT  RIGHT
----  -----
  10     70
  20     60
  30     50
  40   null

Oops :( Misread

with t as ( select level*101 as x from dual connect by level <= 7),
    tt as ( select x, row_number() over (order by x) - 1 as rn, (count(*) over () - 1) / 2 as median_rn from t)
select min(case when rn <= median_rn then x end) as x1,
       min(case when rn >  median_rn then x end) as x2
  from tt
  group by abs(rn - median_rn)
  order by abs(rn - median_rn) desc
;

           X1            X2
------------- -------------
          101           707
          202           606
          303           505
          404

>   group by abs(rn - median_rn)

Another variation of odd count:

  group by abs(trunc(median_rn - rn + 0.1))

           X1            X2
------------- -------------
          101
          202           707
          303           606
          404           505