Oracle SQL to generate interleaved SQL results - sql

Question

Hi i am looking for a way to write a SQL statement which will come out with the following results :-
Lets say we have Dept & Emp ID i would like to generate like records from Dept 3 for the first two rows then followed by Dept 2 with one row then continue Dept 3 and so on :
DEPT EMPID
----- ------
3 1
3 2
2 3
3 7
3 8
2 9
Thank You.

Answer 1

You could use something like this
SELECT
DEPT,
EMPID
FROM (
SELECT
*,
ceil((row_number() OVER (PARTITION BY dept ORDER BY EMPID ))/ 2::numeric(5,2)) AS multiple_row_dept,
row_number() OVER (PARTITION BY dept ORDER BY EMPID ) AS single_row_dept
FROM
test_data2
) sub_query
ORDER BY
CASE
WHEN DEPT = 2 THEN single_row_dept
ELSE multiple_row_dept
END,
DEPT DESC,
EMPID
single_row_dept specifics which dept should appear only once, in this case its DEPT 2 followed by multiple other departments

Answer 2

First order a table by empid in a subquery,
then calculate a remainder of rowids divided by 3,
then depending on a result of calculation, return 2 or 3, using case expression,
like this
SELECT
CASE REMAINDER( rownum, 3 )
WHEN 0 THEN 2
ELSE 3
END As DeptId,
empid
FROM (
SELECT empid
FROM table1
ORDER BY empid
)
demo: http://sqlfiddle.com/#!4/bd1bb/3

Answer 3

The following code has some limitations:
1) There are only 2 DepId's in the source table
2) Ratio between records having different DepId is exactly 2:1
3) Ordering in one place of the query should be changed depending whether DeptId having more records is naturally greater then the other or not
with t as (
select 3 dept_id, 1 emp_id from dual
union all
select 3, 2 from dual
union all
select 3, 7 from dual
union all
select 3, 8 from dual
union all
select 2, 3 from dual
union all
select 2, 9 from dual)
select dept_id, emp_id
from (select dept_id,
emp_id,
dense_rank() over(partition by dept_id order by in_num + mod(in_num, out_num)) ord
from (select dept_id,
emp_id,
row_number() over(partition by dept_id order by emp_id) in_num,
/*in the following ORDER BY change to DESC or ASC depending on which dept_id has more records*/
dense_rank() over(order by dept_id) out_num
from t))
order by ord, dept_id desc;
DEPT_ID EMP_ID
---------- ----------
3 1
3 2
2 3
3 8
3 7
2 9