How to allow my string input to contain only numbers NOT letters

Question

String roulette = keyboard.next();
    if (roulette.length()!=14)
    {
        System.out.print("Error: 14 digits only");
        System.exit(1);
    }

I did this to make sure I have a length of 14. But my program also requires it to be numbers ONLY. It also needs to be a string. I don't want to parse int, I want my program to System.exit(1) if a letter is entered. Is there any way to do this?

I looked here https://docs.oracle.com/javase/8/docs/api/java/util/Scanner.html and https://docs.oracle.com/javase/8/docs/api/java/lang/Integer.html but cant find what im looking for

The character class's has an isDigit method that tells you if a character is a number. It's pretty easy to create your own version of that function though. That would actually be a good exercise for you. Anyways, just use that method to check if every character is a digit. — Carcigenicate
– Carcigenicate, Commented Feb 15, 2017 at 17:12

score 3 · Accepted Answer · 2017-02-15 17:19:14Z

3

You can use a simple regular expression check:

if(roulette.matches("[0-9]{14}")) {
    // has length 14 and only digits
} else {
    // wrong format
}

This checks both for length and if only digits are used.

Further reading on regular expressions: https://docs.oracle.com/javase/tutorial/essential/regex/

edited Feb 15, 2017 at 17:19

answered Feb 15, 2017 at 17:13

user6073886

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2 Comments

Survivorr Over a year ago

Thank you very much, i was breaking my head with isDigit

Survivorr Over a year ago

Any way to make this include ':' too?

Jarvis · Accepted Answer · 2017-02-15 17:24:03Z

0

You use Java regular expressions :

String pattern1 = "\\d{14}";
Pattern p = Pattern.compile(pattern1);
Matcher m = p.matcher(roulette);
//exit if all 14 not integers, or any non-integer present
if(!m.find())
    System.exit(1);

edited Feb 15, 2017 at 17:24

answered Feb 15, 2017 at 17:18

Jarvis

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Comments

Atul · Accepted Answer · 2017-02-15 17:24:35Z

0

Use regex for matching and Scanner for input.

public static void main(String[] args) {
    String regex = "\\d+";
    Scanner s=new Scanner(System.in);
    while(s.hasNext())
    {
        String text=s.next();
        if(text.matches(regex))
        {
            System.out.println("Input Strings only");
        }
        else
        {
            System.out.println(text);
            //break;
        }
    }

answered Feb 15, 2017 at 17:24

Atul

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Comments

Carcigenicate · Accepted Answer · 2017-02-15 21:57:41Z

I'll add this here just to show what I meant by use Character.isDigit. I suggested it since in Clojure, which uses Java, this is easy because it has an every? method which checks if everything in a collection satisfies a predicate (Character.isDigit in this case). In Clojure, this makes it just:

(every? #(Character/isDigit %) "1234")

It's beautiful.

Unfortunately, for a similar method to work in Java, you need to convert the String to a char-array, then ArrayList, then a stream, and call allMatch on it. Instead, here's a manual way using a loop:

import java.util.*;

public class Main {

    public static boolean areAllNums (String numStr) {
        // Loop over each character...
        for (char d : numStr.toCharArray()) {
            //And if we find a non-digit, return false;
            if (!Character.isDigit(d)) {
                return false;
            }
        }
        // If we made it here, all the chars were digits
        return true;
    }

    public static void main(String[] args) {
        System.out.println(Arrays.asList(
            areAllNums("1234"), areAllNums("12g4")));

        // Prints [true, false]
    }
}

mid · Accepted Answer · 2017-02-16 08:15:04Z

0

I found on SO not long ago (Surprised no-one's mentioned yet):

public static boolean isValid(String str) {
    try {
        Double.parseDouble(str);
        return str.length() == 14;
    } catch(NumberFormatException e) {
        return false;
    }
}

answered Feb 16, 2017 at 8:15

mid

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