sql - 1 列のみの列から行へ

Question

テーブルには、すべての個別の値を持つ列が 1 つしかありません。それを 3 つのペアにグループ化し、3 つの行から 3 つの列を作成する必要があります。助けてください

ソース

COL1
-----
A
B
C
D
E
F

必要な出力 1:

COL1
------
A,B,C
D,E,F

必要な出力 2:

col1  col2  col3
----  ----  ----
A     B     C
D     E     F

Answer 1

出力 1:

select listagg(col1, ',') within group (order by col1) as col
from (
  select col1,
         case 
           when row_number() over (order by col1) <= (count(*) over ()) / 2 then 0
           else 1
         end as grp
  from foo
)
group by grp
order by grp;

出力 2 の場合:

select max(col1) as col1,
       max(col2) as col2, 
       max(col3) as col3
from (
  select case mod(row_number() over (order by col1),3)
            when 1 then col1
            else null
         end as col1,
         case mod(row_number() over (order by col1),3)
            when 2 then col1
            else null
         end as col2,         
         case mod(row_number() over (order by col1),3)
            when 0 then col1
            else null
         end as col3,
         case 
           when row_number() over (order by col1) <= (count(*) over ()) / 2 then 0
           else 1
         end as grp
  from foo
)
group by grp
order by grp;

SQLFiddle の例: http://sqlfiddle.com/#!4/d699c/1

Answer 2

別の方法 (レコードの 3 の倍数ごとに機能します)

出力1

select listagg(col1, ',') within group (order by col1) col1
from
(select col1, row_number() over(order by col1) rn
from t) tt
group by rn - decode(mod (rn, 3) ,0,3,mod (rn, 3));

出力2

select c2_col col1, c3_col col2, c1_col col3 
from
(select rn - decode(mod (rn, 3) ,0,3,mod (rn, 3)) grp, mod(rn, 3) rnm, col1
 from
(select col1, row_number() over(order by col1) rn
from t)) tt
pivot
(
  max(col1) as col
  for rnm in (0 as c1,1 c2,2 c3)
  );

ここにsqlfiddleのデモがあります