Java Program To Check String Contains Only Digits (+Java 8 Program)

1. Introduction 

In this article, We will learn how to check whether a string contains only digits or not. This can be done in various ways but covered most possible ways to check the string. If the string is having at least on non-digit or alphabetic or any special character then the function should return false else return true.

The solutionn to this problem relies on digits range in 0 to 9 ASCII Codes, Integer.parseInt(), Character.isDigit(), Java 8 functional style and String.matches() method. Let us start writing example programs for each approach.

Sample Input:

12345

Output:

true

Another Input:

123abc$

Output:

false

2. Integers Range Example (ASCII Codes)

In this approach, We will be running a for loop from index 0 to till string length -1. Next, Get each character and check each char is in the range of 0 to 9. If any of the characters of the string are not in numbers(0 - 9) range then return false. Otherwise, Input string has only digits that return true at last.

Here, we use the ASCII Codes for the numbers range 0 to 9.

0 ASCII code: 49
9 ASCII code: 57

Program:

package com.java.w3schools.blog.java.program.to.strings;

/**
 * 
 * Java program using Numeric and ASCII Codes
 * 
 * @author JavaProgramTo.com
 *
 */
public class StirngOnlyDigitsNumbers {

 public static void main(String[] args) {

  String input = "12345";
  System.out.println("12345 contains only digits : " + checkStringOnlyDigitsWithRange(input));

  System.out.println("999b99 contains only digits : " + checkStringOnlyDigitsWithRange("999b99"));

  System.out.println("$123456789 contains only digits : " + checkStringOnlyDigitsWithRange("$123456789"));

 }

 private static boolean checkStringOnlyDigitsWithRange(String input) {

  for (int i = 0; i < input.length(); i++) {

   int digit = (int) input.charAt(i);

   if (!(digit >= 49 && digit <= 57)) {
    return false;
   }
  }

  return true;
 }

}

Output:

12345 contains only digits : true
999b99 contains only digits : false
$123456789 contains only digits : false

3. Using Integer.parseInt()

Integer.parseInt() method takes string as an argument and converts String into int if the string has only digits. If the string has characters then it will throw NumberFormatException exception.

package com.java.w3schools.blog.java.program.to.strings;

/**
 * 
 * Java program using Integer.parseInt(String input);
 * 
 * @author JavaProgramTo.com
 *
 */
public class StirngOnlyDigitsNumbersParseInt {

 public static void main(String[] args) {

  String input = "12345";
  System.out.println("12345 contains only digits : " + checkStringOnlyDigitParseInt(input));

  System.out.println("999b99 contains only digits : " + checkStringOnlyDigitParseInt("999b99"));

  System.out.println("$123456789 contains only digits : " + checkStringOnlyDigitParseInt("$123456789"));

 }

 private static boolean checkStringOnlyDigitParseInt(String input) {

  try {
   Integer.parseInt(input);
  } catch (NumberFormatException e) {
   return false;
  }

  return true;
 }

}

Output:

12345 contains only digits : true
999b99 contains only digits : false
$123456789 contains only digits : false

4. Using Character.isDigit()

Character.isDigit(char c) method returns true if the provided character is digit else returns false.

/**
 * 
 * Java program using Character.isDigit(ch) method.
 * 
 * @author JavaProgramTo.com
 *
 */
public class StirngOnlyDigitsNumbersIsDigit {

 public static void main(String[] args) {

  String input = "12345";
  System.out.println("12345 contains only digits : " + checkStringOnlyDigitsIsDigit(input));

  System.out.println("999b99 contains only digits : " + checkStringOnlyDigitsIsDigit("999b99"));

  System.out.println("$123456789 contains only digits : " + checkStringOnlyDigitsIsDigit("$123456789"));

 }

 private static boolean checkStringOnlyDigitsIsDigit(String input) {

  for (int i = 0; i < input.length(); i++) {
   char ch = input.charAt(i);
   if (!Character.isDigit(ch)) {
    return false;
   }
  }

  return true;
 }

}

This program also produces the same result.

5. Using String.matches()

String class a matches() method that takes a regular expression and checks the given string matches to the pattern. If matches return true else false.

/**
 * 
 * Java program using Character.isDigit(ch) method.
 * 
 * @author JavaProgramTo.com
 *
 */
public class StirngOnlyDigitsNumbersIsDigit {
 private static final String PATTERN = "[0-9]+";

 private static boolean checkStringOnlyDigitsIsDigit(String input) {

  return input.matches(PATTERN);

 }

}

6. Using Java 8 Streams

Java 8 String class has chars() method that returns IntStream and invokes anyMatches() method.

import java.util.stream.IntStream;

/**
 * 
 * Java program using Java 8 Streams + IntStream
 * 
 * @author JavaProgramTo.com
 *
 */
public class StirngOnlyDigitsNumbersJava8 {

 private static boolean checkStringOnlyDigitsIsDigit(String input) {

  IntStream intStream = input.chars();
  boolean isMatched = intStream.anyMatch(ch -> Character.isDigit(ch));

  return isMatched;

 }

}

7. Conclusion

In this article, We've seen how to check and validate the given string has only digits. Example working programs are shown in various approaches.

Similar, example on count number of digits in a number