REGEXP_REPLACE is matching but not replacing with empty string

Question

Given a table with a column called files_name, and you want to match everything after the match with regex, and replace it with nothing.

I have this code but it matches everything, but it's always changed 0.

UPDATE files 
SET files_name = REGEXP_REPLACE (files_name, 'S[0-9]?[0-9]?[0-9]E[0-9]?[0-9]\K.*', '') 
WHERE type = 5;

A online regex tester says my regex is correct, and is matching as desired, yet no matter what I put into the second '' it never replaces the match.

I'm using mariadb 10.3.17

Answer 1

UPDATE files 
SET files_name =
    REGEXP_REPLACE(
       files_name,
       '(S[0-9]?[0-9]?[0-9]E[0-9]?[0-9])([^0-9].*)',
       '\\1'
    ) 
WHERE type = 5;

fiddle

UPDATE

I executed this command, with redacted info of course...

mariadb -h ip -P port -D database -e "UPDATE files SET files_name = REGEXP_REPLACE (files_name, '(S[0-9]?[0-9]?[0-9]E[0-9]?[0-9])([^0-9].*)', '\\1') WHERE type = 5;" -pPASSWORD

You need to quote a slash:

1. in command line
2. in SQL query

So you would use '\\\\1'.

You may avoid using slash by using CONCAT(CHAR(92), '1') instead of '\\1' in the query text - in that case no slaches in SQL and quoting don't need in.