Few Java examples to show you how to check if a String is numeric.

1. Character.isDigit()

Convert a String into a char array and check it with Character.isDigit()

package com.mkyong;
public class NumericExample {
    public static void main(String[] args) {
        System.out.println(isNumeric(""));          // false
        System.out.println(isNumeric(" "));         // false
        System.out.println(isNumeric(null));        // false
        System.out.println(isNumeric("1,200"));     // false
        System.out.println(isNumeric("1"));         // true
        System.out.println(isNumeric("200"));       // true
        System.out.println(isNumeric("3000.00"));   // false
    public static boolean isNumeric(final String str) {
        // null or empty
        if (str == null || str.length() == 0) {
            return false;
        for (char c : str.toCharArray()) {
            if (!Character.isDigit(c)) {
                return false;
        return true;

Output

false
false
false
false
true
true
false

2. Java 8

This is much simpler now.

	public static boolean isNumeric(final String str) {
        // null or empty
        if (str == null || str.length() == 0) {
            return false;
        return str.chars().allMatch(Character::isDigit);

3. Apache Commons Lang

If Apache Commons Lang is present in the claspath, try NumberUtils.isDigits()

	<dependency>
		<groupId>org.apache.commons</groupId>
		<artifactId>commons-lang3</artifactId>
		<version>3.9</version>
	</dependency>

import org.apache.commons.lang3.math.NumberUtils;
	public static boolean isNumeric(final String str) {
        return NumberUtils.isDigits(str);

4. NumberFormatException

This solution is working, but not recommend, performance issue.

	public static boolean isNumeric(final String str) {
        if (str == null || str.length() == 0) {
            return false;
        try {
            Integer.parseInt(str);
            return true;
        } catch (NumberFormatException e) {
            return false;