Find each position of same characters in a string through SQL query - sql

Question

Below is my data set
Year Month Holiday list
----------------------------------------------------------
2007 1 WWHHWWWWWHHWWWWWHHWWWWWHHWWWWWH
2008 4 HWWWWWHHWWWWWHHWWWWWHHWWWWWHHWW
I want to write SQL query to get below output wherein the position of each "H" is displayed.
Year Month Holiday list
-----------------------------------------------------
2007 1 3 4 10 11 18 19 25 26 31
2008 4 1 7 8 14 15 21 22 28 29
The INSTR function in Oracle returns the position of first character. So, I cannot use it.
I could write a Function and pass the Holiday List column value to it. Loop through the string and
find the position of each "H". But, I want to achieve this through plain SQL select query.
How to find each position of same characters in a string through SQL query in Oracle database?

Answer 1

Something like this might do it. You can use the INSTRfunction and a hierarchical query to go through the string and then use the LISTAGG analytic function to concatenate the results.
WITH data AS (
SELECT 2007 year, 1 month, 'WWHHWWWWWHHWWWWWHHWWWWWHHWWWWWH' holiday_list FROM DUAL
UNION
SELECT 2008 year, 4 month, 'HWWWWWHHWWWWWHHWWWWWHHWWWWWHHWW' holiday_list FROM DUAL)
SELECT year, month, (SELECT LISTAGG(INSTR(holiday_list,'H',1,LEVEL),' ') WITHIN GROUP (ORDER BY LEVEL)
FROM DUAL
CONNECT BY LEVEL < INSTR(holiday_list,'H',1,LEVEL)) S
FROM data;

Answer 2

Well you can get a list, see the dbfiddle here
SELECT id, INSTR(some, 'H', 1, c.no) AS Pos
FROM tst CROSS JOIN consecutive c
WHERE INSTR(some, 'H', 1, c.no) > 0
This gives you a list of the days with an H. You could then try to transpose this list. But be warned, there is an intentional CROSS JOIN there, so for every row in your data you will get up to 31 rows with this statement. Like Tim Biegeleisen said, it may be better to use an UDF for this.