Find and Replace in SQLite being mucked up by percent sign.

Question

I'm trying to use this code:

UPDATE media_items SET content_url = replace(content_url,’cd%20images’,'my%20music’);

And I know that the percent signs are mucking it up, and I have to use some sort of escape to get them to read properly, but I've done a bunch of searching, and anything I find doesn't seem to work. What is the proper syntax to include an escape for the percent signs to get it to work correctly?

Answer 1

I think you just double them: replace(content_url, "cd%%20images", "my%%20music").