CODEWITHSUNDEEP
javaregex
haedes
haedes

Reputation: 622

Matching the pattern of a string in java

I have been trying to figure out how to match the pattern of my input string with this kind of string:

"xyz 123456789"

In general every time I have a input that has first 3 characters (can be both uppercase or lowercase) and last 9 are digits (any combination) the input string should be accepted.

So if I have i/p string = "Abc 234646593" it should be a match (one or two white-space allowed). Also it would be great if "Abc" and "234646593" should be stored in seperate strings.

I have seeing a lot of regex but do not fully understand it.

Upvotes: 1

Views: 939

Answers (1)

quux00
quux00

Reputation: 14624

Here's a working Java solution:

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Regex {
  public static void main(String[] args) {
    String input = "Abc 234646593";

    // you could use \\s+ rather than \\s{1,2} if you only care that
    // at least one whitespace char occurs
    Pattern p = Pattern.compile("([a-zA-Z]{3})\\s{1,2}([0-9]{9})");
    Matcher m = p.matcher(input);
    String firstPart = null;
    String secondPart = null;
    if (m.matches()) {
      firstPart = m.group(1);  // grab first remembered match ([a-zA-Z]{3})
      secondPart = m.group(2); // grab second remembered match ([0-9]{9})
      System.out.println("First part: " + firstPart);
      System.out.println("Second part: " + secondPart);
    }
  }
}

Prints out:

First part: Abc
Second part: 234646593

Upvotes: 4

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