Oracle Regex Connect By - sql

Question

I am trying to produce multiple rows after performing a regex on a column splitting all values in square brackets. I'm only able to return a single value though, currently.
The field I am performing the regex has this value:
[1265]*[1263]
I am trying to get 1265 and 1263 in my result set as separate rows.
SELECT REGEXP_SUBSTR(column,'\[(.*?)\]',1,LEVEL) AS "col1"
FROM table
CONNECT BY REGEXP_SUBSTR(column,'\[(.*?)\]',1,LEVEL) IS NOT NULL;
Instead I just get this in the result set.
[1263]

Answer 1

with test (rn, col) as
(
select 'a', '[123]*[abc] []' from dual union all
select 'b', '[45][def] ' from dual union all
select 'c', '[678],.*' from dual
),
coll (rn, col) as
(
select rn,regexp_replace(col, '(\[.*?\])|.', '\1') from test
),
cte (rn, cnt, col, i) as
(
select rn, 1, col, regexp_substr(col, '(\[(.*?)\])', 1, 1, null, 2)
from coll
union all
select rn, cnt+1, col, regexp_substr(col, '(\[(.*?)\])', 1, cnt+1, null, 2)
from cte
where cnt+1 <= regexp_count(col, '\[.*?\]')
)
select * from cte
order by 1,2;

Answer 2

This regex counts elements by looking for closing brackets and returns the digits inside the brackets, allowing for NULLs. Separators are ignored since the data elements you want are surrounded by square brackets we can focus on those.
SQL> with test(rownbr, col) as (
select 1, '[1265]**[1263]' from dual union
select 2, '[123]' from dual union
select 3, '[111][222]*[333]' from dual union
select 4, '[411]*[][433]' from dual
)
select distinct rownbr, level as element,
regexp_substr(col, '\[([0-9]*)\]', 1, level, null, 1) value
from test
connect by level <= regexp_count(col, ']')
order by rownbr, element;
ROWNBR ELEMENT VALUE
---------- ---------- -----
1 1 1265
1 2 1263
2 1 123
3 1 111
3 2 222
3 3 333
4 1 411
4 2
4 3 433
9 rows selected.
SQL>