Question

I’m trying to find ‘%%’ string in content.

SELECT * FROM myTable WHERE myColumn LIKE '%\%\%%'

But it returns also the rows, where only one percent sign found.

How to find only the rows, where two percent signs exists?

Thanks.

Answer 1

If you are indeed only looking for literal % and don’t need the wildcards, this would be made simpler using LOCATE(), where the % will be parsed as literals requiring no escaping.

SELECT * FROM myTable WHERE LOCATE('%%', myColumn) > 0;