Explain Regular Expression \s Metacharacter in Java

The subexpression/metacharacter “\s” matches the white space equivalent.

Example 1

import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegexExample {
   public static void main( String args[] ) {
      String regex = "\\s";
      String input = "Hello how are you welcome to Tutorialspoint !";
      Pattern p = Pattern.compile(regex);
      Matcher m = p.matcher(input);
      int count = 0;
      while(m.find()) {
         count++;
      }
      System.out.println("Number of matches: "+count);
   }
}

Output

Number of matches: 7

Example 2

The following example reads a string and removes all the extra spaces between them.

import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Example {
   public static void main(String args[]) {
      //Reading String from user
      System.out.println("Enter a String");
      Scanner sc = new Scanner(System.in);
      String input = sc.nextLine();
      //Regular expression to match spaces (one or more)
      String regex = "\\s+";
      //Compiling the regular expression
      Pattern pattern = Pattern.compile(regex);
      //Retrieving the matcher object
      Matcher matcher = pattern.matcher(input);
      //Replacing all space characters with single space
      String result = matcher.replaceAll(" ");
      System.out.print("Text after removing unwanted spaces: \n"+result);
   }
}

Output

Enter a String
hello this is a sample text with irregular spaces
Text after removing unwanted spaces:
hello this is a sample text with irregular spaces