Question

I have a seemingly simple problem though i am unable to get my head around it.

Let’s say i have the following string: ‘abcabcabcabc’ and i want to get the last occurrence of ‘ab’. Is there a way i can do this without looping through all the other ‘ab’s from the beginning of the string?

I read about anchoring the end of the string and then parsing the string with the required regular expression. I am unsure how to do this in Java (is it supported?).

Update: I guess i have caused a lot of confusion with my (over) simplified example. Let me try another one. Say, i have a string as thus – ’12/08/2008 some_text 21/10/2008 some_more_text 15/12/2008 and_finally_some_more’. Here, i want the last date and hence i need to use regular expressions. I hope this is a better example.

Thanks, Anirudh

Answer 1

Firstly, thanks for all the answers.

Here is what i tried and this worked for me:

Pattern pattern = Pattern.compile('(ab)(?!.*ab)'); Matcher matcher = pattern.matcher('abcabcabcd'); if(matcher.find()) {   System.out.println(matcher.start() + ', ' + matcher.end()); } 

This displays the following:

6, 8 

So, to generalize – <reg_ex>(?!.*<reg_ex>) should solve this problem where ‘?!’ signifies that the string following it should not be present after the string that precedes ‘?!’.

Update: This page provides a more information on ‘not followed by’ using regex.