regex - match punctuation at end of word in Java/Scala String

Question

I have this regex example at this link: http://regexr.com/39rr0

Text input:
1. This camera support Monochrome, Neutral, Standard, Landscape and Portrait!

Expected Output on string replace using this regex:
1 This camera support Monochrome Neutral Standard Landscape and Portrait

I'm trying to remove punctuation which comes after words so I get only the words without it. My regex is (([\\S]+)(,|:|;|\\?|!)) and it matches fine in that regex editor. However when I do string.replace( (([\\S]+)(,|:|;|\\?|!)), "") or even string.find( (([\\S]+)(,|:|;|\\?|!)) ) it finds nothing.

What's the regex to do this? Is my regex broken or am I using it incorrectly.

Answer 1

You cannot use String.replace , it only replaces literals.

Try a simpler Pattern and String.replaceAll :

System.out.println(
    "1. This camera support Monochrome, Neutral, Standard, Landscape and Portrait!"
    .replaceAll("\\p{Punct}", "")
);

Output

1 This camera support Monochrome Neutral Standard Landscape and Portrait

Note

If you only need to replace punctuation after a word character (an alphanumeric), you can improve the Pattern as such:

"(?<=\\w)\\p{Punct}+"

In your case, it will produce the same output.

Answer 2

Use \\\\p{P} to match all the punctuations.

String str = "1. This camera support Monochrome, Neutral, Standard, Landscape and Portrait!";
System.out.println(str.replaceAll("\\p{P}(?=\\s|$)", ""));

Output:

1 This camera support Monochrome Neutral Standard Landscape and Portrait

Explanation:

• \\\\p{P} Match all the punctuations
• (?=\\\\s|$) Only if it's followed by a space or end of the line anchor.

Answer 3

Use replaceAll, it works with regexp

System.out.println("This camera support Monochrome, Neutral, Standard, Landscape and Portrait!".replaceAll("([\\S]+)(,|:|;|\\?|!)", "$1"));

Prints :

This camera support Monochrome Neutral Standard Landscape and Portrait