Regular expression to match optional end of string

Tags: java regex

Given the following:

"John Smith"
"John Smith (123)"
"John Smith (123) (456)"

I'd like to capture:

"John Smith"
"John Smith", "123"
"John Smith (123)", "456"

What Java regex would allow me to do that?

I've tried (.+)\s\((\d+)\)$ and it works fine for "John Smith (123)" and "John Smith (123) (456)" but not for "John Smith". How can I change the regex to work for the first input as well?


You may turn the first .+ lazy, and wrap the later part with a non-capturing optional group:

   ^ ^^^           ^^ 

See the regex demo

Actually, if you are using the regex with String#matches() the last $ is redundant.


  • (.+?) - Group 1 capturing one or zero characters other than a linebreak symbol, as few as possible (thus, allowing the subsequent subpattern to "fall" into a group)
  • (?:\s\((\d+)\))? - an optional sequence of a whitespace, (, Group 2 capturing 1+ digits and a )
  • $ - end of string anchor.

A Java demo:

String[] lst = new String[] {"John Smith","John Smith (123)","John Smith (123) (456)"};
Pattern p = Pattern.compile("(.+?)(?:\\s\\((\\d+)\\))?");
for (String s: lst) {
    Matcher m = p.matcher(s);
    if (m.matches()) {
        if ( != null)

try something like this.

You need the variable accessible in both functions so it needs to be declared outside those functions and then defined inside the first one and use clearInterval() to cancel an existing interval

var timer;
$('#name').mouseenter(function() {
   timer = setInterval(mosaic, 150);
}).mouseleave(function() {

This video can help you solving your question :)
By: admin