Remove begining punctuation from a word

Question

I have seen a couple of threads here that kindof matches what I am asking here. But none are concrete. If I have a string like "New Delhi", I want my code to extract New Delhi. So here the quotes are stripped off. I want to strip off any punctuation, in general at start and end.

So far, this helps to strip the punctuations at the end:

String replacedString = replaceable_string.replaceAll("\\p{Punct}*([a-z]+)\\p{Punct}*", "$1");

What am I doing wrong here? My output is "New Delhi with the beginning quote still there.

Answer 1

The following will remove a punctuation character from both the beginning and end of a String object if present:

String s = "\"New, Delhi\"";

// Output: New, Delhi
System.out.println(s.replaceAll("^\\p{Punct}|\\p{Punct}$", ""));

The ^ part of the Regex represents the beginning of the text, and $ represents the end of the text. So, ^\p{Punct} will match a punctuation that is a first character and \p{Punct}$ will match a punctuation that is a last character. I used | (OR) to match either the first expression or the second one, resulting in ^\p{Punct}|\p{Punct}$.

In case you want to remove all punctuation characters from the beginning and the end of the String object, you can use the following:

String s = "\"[{New, Delhi}]\"";

// Output: New, Delhi
System.out.println(s.replaceAll("^\\p{Punct}+|\\p{Punct}+$", ""));

I simply added the + sign after each \p{Punct}. The + sign means "One or more", so it will match many punctuations if they are present at the beginning or end of the text.

Hope this is what you were looking for :)