On Monday, June 21, 2021, Oliver Kohll <
oliver@agilechilli.com> wrote:
select regexp_replace(
'here is [[my text]] to replace and [[some more]]',
E'\\[\\[(.*?)\\]\\]',
replace(E'\\1', ' ', '_'),
'g'
);
Side note, you seldom want to use “E” (escape) string literals with regexes (or in general really) using just the simple literal syntax removes the need for double-backslashing.
David J.