Split a column into multiple rows - oracle

Question

I have data like this my table:
Year
Month
Daytype
"2020"
"01"
"HBBBBHHBBBBBHHBBBBBHHBBBBBHHBBB"
"2020"
"02"
"BBHHBBBBBHHBBBBBHHBBBBBHHBBB"
Now I need to convert this table as below ;
Year
Month
Day
Daytype
2020
01
01
H
2020
01
02
B
2020
01
03
B
2020
01
04
B
2020
01
05
B
2020
01
06
H
2020
01
07
H
I tried option using regular expression. but my problem the daytype string doesn't have any delimiter string.
Can someone assist me. I want this in SQL only

Answer 1

You can use a recursive query and simple string functions (which are much faster than regular expressions):
WITH days (year, month, daytypes, day, daytype) AS (
SELECT year, month, daytype, 1, SUBSTR(daytype, 1, 1)
FROM table_name
UNION ALL
SELECT year, month, daytypes, day + 1, SUBSTR(daytypes, day + 1, 1)
FROM days
WHERE day < LENGTH(daytypes)
)
SELECT year, month, day, daytype
FROM days
ORDER BY year, month, day;
Which, for the sample data:
CREATE TABLE table_name (Year,Month,Daytype) AS
SELECT 2020, 1, 'HBBBBHHBBBBBHHBBBBBHHBBBBBHHBBB' FROM DUAL UNION ALL
SELECT 2020, 2, 'BBHHBBBBBHHBBBBBHHBBBBBHHBBB' FROM DUAL;
Outputs:
YEAR
MONTH
DAY
DAYTYPE
2020
1
1
H
2020
1
2
B
2020
1
3
B
2020
1
4
B
2020
1
5
B
2020
1
6
H
2020
1
7
H
2020
1
8
B
2020
1
9
B
2020
1
10
B
2020
1
11
B
2020
1
12
B
2020
1
13
H
2020
1
14
H
2020
1
15
B
2020
1
16
B
2020
1
17
B
2020
1
18
B
2020
1
19
B
2020
1
20
H
2020
1
21
H
2020
1
22
B
2020
1
23
B
2020
1
24
B
2020
1
25
B
2020
1
26
B
2020
1
27
H
2020
1
28
H
2020
1
29
B
2020
1
30
B
2020
1
31
B
2020
2
1
B
2020
2
2
B
2020
2
3
H
2020
2
4
H
2020
2
5
B
2020
2
6
B
2020
2
7
B
2020
2
8
B
2020
2
9
B
2020
2
10
H
2020
2
11
H
2020
2
12
B
2020
2
13
B
2020
2
14
B
2020
2
15
B
2020
2
16
B
2020
2
17
H
2020
2
18
H
2020
2
19
B
2020
2
20
B
2020
2
21
B
2020
2
22
B
2020
2
23
B
2020
2
24
H
2020
2
25
H
2020
2
26
B
2020
2
27
B
2020
2
28
B
fiddle