Generate numbers for given ranges (sql query) - sql

Question

I'm writing a query which generate numbers between specified ranges.
I have a table
NUM_RANGES
ID START_NUMBER END_NUMBER
-- ------------ ----------
1 1 5
2 6 10
I need to get this result:
ID NUMBER
-- ------
1 1
1 2
1 3
1 4
1 5
2 6
2 7
2 8
2 9
2 10
With this query i can get correct result but only for specified id in where clause:
select id, start_number + level - 1 next_tag
from (select id, start_number,end_number
from NUM_RANGES
where id = 1
)
connect by level <= end_number - start_number + 1
Without "where id = 1" im getting 62 rows with duplicates where distinct helps but with bigger ranges 1 - 200 , 200- 500 it works too slow ..
Thanks for help!

Answer 1

On Oracle 12c you can use CROSS APPLY:
select *
from num_ranges
cross apply(
select level - 1 + start_number as my_new_number
from dual
connect by level <= end_number - start_number + 1
);

Answer 2

In Oracle 11.2 and earlier, you can use a hierarchical query:
with
num_ranges ( id, start_number, end_number ) as (
select 1, 1, 5 from dual union all
select 2, 9, 12 from dual
)
-- End of simulated input data (for testing purposes only, not part of the solution).
-- SQL query begins below this line.
select id, start_number + level - 1 as nmbr
from num_ranges
connect by level <= end_number - start_number + 1
and prior id = id
and prior sys_guid() is not null
order by id, nmbr -- If needed
;
ID NMBR
---------- ----------
1 1
1 2
1 3
1 4
1 5
2 9
2 10
2 11
2 12
Specifically, without linking new rows to existing ones by the same id, you are generating an insane number of unnecessary rows. This is why you need prior id = id. The additional condition, prior sys_guid() is not null, is needed so that Oracle does not see cycles where it shouldn't see them (which are caused precisely by "prior id = id").
In Oracle 12.1 or higher, you can use the lateral clause:
select n.id, l.nmbr
from num_ranges n,
lateral ( select start_number + level - 1 as nmbr from dual
connect by level <= end_number - start_number + 1 ) l
order by id, nmbr -- If needed
;

Answer 3

This should work. Pretty fast too.
with cte as (select 0 as c from dual)
, cte4 as (select c from cte union all select c from cte union all select c from cte union all select c from cte)
, cte256 as (select t0.c from cte4 t0, cte4 t1, cte4 t2, cte4 t3)
, nums as (select row_number() over(order by null) as n from cte256 t0, cte256 t1, cte256 t2)
select NR.id, nums.n as NUMBER_
from nums
join NUM_RANGES NR on nums.n between NR.START_NUMBER and NR.END_NUMBER
;