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