Replace all CRLF with \r\n using REXEXP_REPLACE - sql

Question

I have a process that needs to strip all CRLF from SQL statements and store them with literal characters \r\n and stores this in a .json file. Later, this process is reversed when the .json file is inserted back into the database.
I am using ORACLE 19c.
The original (abbreviated) SQL might look like this:
SELECT SPRIDEN_ID, SPRIDEN_LAST_NAME, SPRIDEN_FIRST_NAME
FROM sztdhof
LEFT JOIN SPRIDEN
ON sztdhof.SZTDHOF_PIDM = spriden.SPRIDEN_PIDM
RIGHT JOIN SSBSECT
ON sztdhof_CRN = SSBSECT_CRN
WHERE spriden_change_ind IS NULL
The resulting string should look like this:
SELECT SPRIDEN_ID, SPRIDEN_LAST_NAME, SPRIDEN_FIRST_NAME\r\nFROM sztdhof\r\nLEFT JOIN SPRIDEN\r\nON sztdhof.SZTDHOF_PIDM = spriden.SPRIDEN_PIDM\r\nRIGHT JOIN SSBSECT\r\nON sztdhof_CRN = SSBSECT_CRN \r\nWHERE spriden_change_ind IS NULL
I have tried numerous ways to get ORACLE REGEXP_REPLACE to recognize the carriage return and line feed together. None seem to work.
To start, I read the value from the db into a variable v_text. Then I attempt the conversion and placing the result into a second variable v_outtext. If I use two separate statements, each works separately. And I can live with that method, but I would like this to work as a single statement. (Note. I have also tried the REPLACE function with similar results.)
This code...
SELECT REPLACE (v_text, chr(13) , '\r\n')
INTO v_outtext
FROM dual;
produces...
SELECT SPRIDEN_ID, SPRIDEN_LAST_NAME, SPRIDEN_FIRST_NAME\r\n
FROM sztdhof\r\n
LEFT JOIN SPRIDEN\r\n
ON sztdhof.SZTDHOF_PIDM = spriden.SPRIDEN_PIDM\r\n
RIGHT JOIN SSBSECT\r\n
ON sztdhof_CRN = SSBSECT_CRN \r\n
WHERE spriden_change_ind IS NULL\r\n
SELECT REGEXP_REPLACE(v_text, chr(10) , '\r\n') --is similar to above, except the \r\n are at the front of each line.
SELECT REGEXP_REPLACE(v_text, (chr(13)chr(10) , '\r\n') gives me an error, missing right parenthesis.
SELECT REGEXP_REPLACE (v_text, '(\r\n)', '\r\n') replaces r with \r and n with \n, but only the first instance.
I have tried numerous iterations. None solve the problem. Any thoughts would be welcome.

Answer 1

You need to concatenate the two characters you want to replace, not just put them in parentheses (even if you supplied both):
SELECT REPLACE(v_text, chr(13) || chr(10) , '\r\n')
INTO v_outtext
FROM dual;
If if it's already in a PL/SQL variable you don't need to to the context switch to select from dual; you can simplify to:
v_outtext := REPLACE(v_text, chr(13) || chr(10) , '\r\n');
but it sounds like you can use the first method to read the value from the database into v_outtext in one go, without the intermediate v_text.
db<>fiddle