CODEWITHSUNDEEP
javaregex
Danny Delott
Danny Delott

Reputation: 7008

Regex to match continuous pattern of integer then space

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 enumerated list of items, and they type in the numbers for the items they want to select, separated by a space. Here is an example:

1   yorkshire terrier
2   staffordshire terrier
3   goldfish
4   basset hound
5   hippopotamus 

Type the numbers that correspond to the words you wish to exclude: 3 5

The enumerated list of items can be a just a few elements or several hundred. The current regex I'm using looks like this ^|\\.\\s+)\\d+\\s+, but I know it's wrong. I don't fully understand regular expressions yet, so if you can explain what it is doing that would be helpful too!

Upvotes: 2

Views: 6671

Answers (4)

Valhure
Valhure

Reputation: 76

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.

Upvotes: 6

Casimir et Hippolyte
Casimir et Hippolyte

Reputation: 89584

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.

Upvotes: 0

Bohemian
Bohemian

Reputation: 425208

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));

Upvotes: 0

Joey
Joey

Reputation: 354714

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 \\.

Upvotes: 2

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