Simple validation example

In this example we shall show you how to make a simple validation of a String, using a Matcher against a specified Pattern. To make a simple String validation one should perform the following steps:

  • Create a new Pattern, by compiling to it a regular expression. The regular expression constructed here is the word “Java” followed by a space character and one digit. In order to do so, use compile(String regex) API method of Pattern.
  • Use matcher(CharSequence input) API method of Pattern to create a Matcher that will match the given String input against this pattern.
  • Use find() API method of Matcher to try to find the next subsequence of the String input sequence that matches the pattern. The method returns true if, and only if, a subsequence of the input sequence matches this matcher’s pattern. In the example the given in input contains the word “Java” followed by a space character and the digit 5, so the method returns true,

as described in the code snippet below.
Note that the compile(String regex) API method of Pattern might throw a PatternSyntaxException, that indicates a syntax error in the regular-expression pattern. The application exits if this exception occurs, with System.exit(0) API method.

package com.javacodegeeks.snippets.core;

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

public class Main {

    public static void main(String args[]) {

  Pattern pattern = null;

  try {

pattern = Pattern.compile("Java \\d");

  } catch (PatternSyntaxException e) {




  String str = "I love Java 5";

  Matcher m = pattern.matcher(str);

  System.out.println("result=" + m.find());



This was an example of how to make a simple validation of a String, using a Matcher against a Pattern in Java.

Ilias Tsagklis

Ilias is a software developer turned online entrepreneur. He is co-founder and Executive Editor at Java Code Geeks.
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