Regex to match continuous pattern of integer then space

Question

I\'m asking the user for input through the Scanner in Java, and now I want to parse out their selections using a regular expression. In essence, I show them an enu

Answer 1

To "parse out" the integers, you don't necessarily want to match the input, but rather you want to split it on spaces (which uses regex):

String[] nums = input.trim().split("\\s+");

If you actually want int values:

List<Integer> selections = new ArrayList<>();
for (String num : input.trim().split("\\s+"))
    selections.add(Integer.parseInt(num));

Answer 2

If you want to ensure that your string contains only numbers and spaces (with a variable number of spaces and trailing/leading spaces allowed) and extract number at the same time, you can use the \G anchor to find consecutive matches.

String source = "1 3 5 8";

List<String> result = new ArrayList<String>();
Pattern p = Pattern.compile("\\G *(\\d++) *(?=[\\d ]*$)");
Matcher m = p.matcher(source);

while (m.find()) {
    result.add(m.group(1));
}
for (int i=0;i<result.size();i++) {
    System.out.println(result.get(i));
}

Note: at the begining of a global search, \G matches the start of the string.

Answer 3

You want integers:

\d+

followed by any number of space, then another integer:

\d+( \d+)*

Note that if you want to use a regex in a Java string you need to escape every \ as \\.

Answer 4

Pattern pattern = new Pattern(^([0-9]*\s+)*[0-9]*$)

Explanation of the RegEx:

• ^ : beginning of input
• [0-9] : only digits
• '*' : any number of digits
• \s : a space
• '+' : at least one space
• '()*' : any number of this digit space combination
• $: end of input

This treats all of the following inputs as valid:

• "1"
• "123 22"
• "123 23"
• "123456 33 333 3333 "
• "12321 44 452 23 "

etc.