Regex to find dd/mm/yyyy

Question

I have a pretty simple code and in one case it works, and not in all the case, I wanted to know what I'm missing I'm trying to match a date in this format dd/mm/yyyy

This is my code:

String d = '20/03/2020';
System.debug(Pattern.matches('(0[1-9]|1[012])/(0[1-9]|[12][0-9]|3[01])/([0-9]{4})', d)); // returns false
d.replace('(0[1-9]|1[012])/(0[1-9]|[12][0-9]|3[01])/([0-9]{4})', 'test');
System.debug(d); //returns D inchanged : 20/03/2020 and not 'test'

I also tried to simply the regex and there the Pattern function works...

String d = '20/03/2020';
System.debug(Pattern.matches('[0-9]{2}/[0-9]{2}/[0-9]{4}', d)); // returns true
d.replace('([0-9]{2}/[0-9]{2}/[0-9]{4})', 'test');
System.debug(d); //returns d inchanged : 20/03/2020 and not 'test'

I've tried many many many solution and regexs and I can't figure out why in a particular case it seems to works

Thanks

Answer 1

There are two issues:

1. replace(String target, String replacement) doesn't take a regExp as input, but a literal string.
   You should have used replaceAll(String regExp, String replacement)
2. Strings are immutable, so doing d.replace() or d.replaceAll() will not change that instance. Indeed they return a new string, so you should assign it to d.

Therefore the code should look like:

String d = '20/03/2020';
System.debug(Pattern.matches('[0-9]{2}/[0-9]{2}/[0-9]{4}', d)); // true
d = d.replaceAll('[0-9]{2}/[0-9]{2}/[0-9]{4}', 'test');
System.debug(d); // test