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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
Updated on: 19-Nov-2019
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