How to check if a character in a string is a digit or letter

Question

I have the user entering a single character into the program and it is stored as a string. I would like to know how I could check to see if the character that was entered is a letter or a digit. I have an if statement, so if its a letter its prints that it's a letter, and the same for a digit. The code I have so far doesn't work but I feel like I'm close. Any help you can offer is appreciated.

  System.out.println("Please enter a single character: ");
  String character = in.next();

  System.out.println(character);

  if (character.isLetter()){
    System.out.println("The character entered is a letter.");
  }
  else (character.isDigit()){
    Syste.out.println("The character entered is a digit.");

the else statement has two compile errors, insert assignment operator expression. — user1701604, Oct 03 '12 at 19:16
You name a String variable "character" then you call the method isLetter() which doesn't exist in the String class, but does exist in the Character class, but "character" is not a Character. The question is very tricky. — Tulains Córdova, Oct 03 '12 at 19:23

score 76 · Answer 1 · answered Oct 03 '12 at 19:15

76

You could use:

    if (Character.isLetter(character.charAt(0))){
    ....

answered Oct 03 '12 at 19:15

Reimeus

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it says this method is undefined – user1701604 Oct 03 '12 at 19:18
See the [docs](http://docs.oracle.com/javase/7/docs/api/java/lang/Character.html#isLetter%28char%29) – Reimeus Oct 03 '12 at 19:20
@Reimeus - Do variables of type `char` have a `.charAt(0)` method? – Richard JP Le Guen Oct 03 '12 at 19:21
1

No. `character` is actually a string in the question. – Reimeus Oct 03 '12 at 19:22
1

the problem coming back is at String character = in.next(). It says "in cannot be resolved" – user1701604 Oct 03 '12 at 19:23
Nvm, fixed it, forgot to do Scanner in = new Scanner(System.in); – user1701604 Oct 03 '12 at 19:24
How would I apply this to an array? – Jonco98 Nov 16 '15 at 20:49
It's unclear what you're asking. If you put all the details in a new question, there will be many people only too delighted to answer :) – Reimeus Nov 16 '15 at 21:12

Hristo · Answer 2 · 2012-10-03T19:22:47.253

You could use the existing methods from the Character class. Take a look at the docs:

http://download.java.net/jdk7/archive/b123/docs/api/java/lang/Character.html#isDigit(char)

So, you could do something like this...

String character = in.next();
char c = character.charAt(0);
...
if (Character.isDigit(c)) { 
    ... 
} else if (Character.isLetter(c)) {
    ...
}
...

If you ever want to know exactly how this is implemented, you could always look at the Java source code.

score 16 · Answer 3 · edited Sep 09 '16 at 15:29

16

Ummm, you guys are forgetting the Character.isLetterOrDigit method:

boolean x;
String character = in.next();
char c = character.charAt(0);
if(Character.isLetterOrDigit(charAt(c)))
{ 
  x = true;
}

edited Sep 09 '16 at 15:29

Nick Roth

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answered Oct 27 '14 at 23:43

Fluffle

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1

I think it's `isLetterOrDigit()` – Raven Dec 21 '15 at 19:57
2

This doesn't tell you which of the two it is. It only verifies that it is at least a number or digit. It would return true for "A" or "2". – C LaBelle Dec 06 '16 at 16:11

score 7 · Answer 4 · edited Jul 07 '18 at 11:51

This is a little tricky, the value you enter at keyboard, is a String value, so you have to pitch the first character with method line.chartAt(0) where, 0 is the index of the first character, and store this value in a char variable as in char c= line.charAt(0) now with the use of method isDigit() and isLetter() from class Character you can differentiate between a Digit and Letter.

here is a code for your program:

import java.util.Scanner;

class Practice
{
 public static void main(String[] args)
  {
   Scanner in = new Scanner(System.in);
   System.out.println("Input a letter"); 
   String line = in.nextLine();
   char c = line.charAt(0);
   if( Character.isDigit(c))
   System.out.println(c +" Is a digit");
   else if (Character.isLetter(c))
   System.out.println(c +" Is a Letter");
  }

}

score 6 · Answer 5 · answered Oct 03 '12 at 19:21

6

By using regular expressions:

boolean isChar = character.matches("[a-zA-z]{1}");
boolean isDigit = character.matches("\\d{1}");

answered Oct 03 '12 at 19:21

Michael

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score 4 · Answer 6 · edited Oct 03 '12 at 19:23

4

char charInt=character.charAt(0);   
if(charInt>=48 && charInt<=57){
    System.out.println("not character");
}
else
    System.out.println("Character");

Look for ASCII table to see how the int value are hardcoded .

edited Oct 03 '12 at 19:23

Baz

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answered Oct 03 '12 at 19:23

Jimmy

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@vandey.. Why would you want to re-invent the wheel?? There already pre-defined method to do this task for you.. – Rohit Jain Oct 03 '12 at 19:26
1

It is just that if the OP wants to learn, he can understand what actually is going on when he calls those method. – Jimmy Oct 03 '12 at 19:29
@RohitJain, `Character.isDigit` cheks for digit in **any** digit system, and this solution will ignore anything except arabic digits – msangel May 30 '14 at 16:52

score 3 · Answer 7 · edited Jun 19 '18 at 06:14

3

This is the way how to check whether a given character is alphabet or not

public static void main(String[] args) {

    Scanner sc = new Scanner(System.in);
    char c = sc.next().charAt(0); 
    if((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z'))
         System.out.println(c + " is an alphabet.");
    else
        System.out.println(c + " is not an alphabet.");
}

edited Jun 19 '18 at 06:14

Rajat Jain

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answered Jun 19 '18 at 05:22

malith vitha

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What if it's special symbol ? – Amol Bais Apr 20 '21 at 13:24

score 2 · Answer 8 · answered Oct 03 '12 at 19:16

2

     char temp = yourString.charAt(0);
     if(Character.isDigit(temp))
     {
         ..........
     }else if (Character.isLetter(temp))
     {
          ......
      }else
     {
      ....
     }

answered Oct 03 '12 at 19:16

kosa

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score 2 · Answer 9 · edited Mar 15 '17 at 11:34

2

import java.util.*;

public class String_char 
{

    public static void main(String arg[]){   

        Scanner in = new Scanner(System.in);
        System.out.println("Enter the value");
        String data;
        data = in.next();

        int len = data.length();
        for (int i = 0 ; i < len ; i++){
            char ch = data.charAt(i);

            if ((ch >= '0' && ch <= '9')){
                System.out.println("Number ");
            }
            else if((ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z')){
            System.out.println("Character");
            }
            else{
                System.out.println("Symbol");
             }
        }
    } 
}

edited Mar 15 '17 at 11:34

KarelG

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answered Mar 15 '17 at 11:17

Jain Austin

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While this code snippet may solve the question, [including an explanation](http://meta.stackexchange.com/questions/114762/explaining-entirely-code-based-answers) really helps to improve the quality of your post. Remember that you are answering the question for readers in the future, and those people might not know the reasons for your code suggestion. – DimaSan Mar 15 '17 at 11:26

score 1 · Answer 10 · answered Oct 03 '12 at 19:17

1

You need to convert your string into character..

String character = in.next();
char myChar = character.charAt(0);

if (Character.isDigit(myChar)) {
   // print true
}

Check Character for other methods..

answered Oct 03 '12 at 19:17

Rohit Jain

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Jeffy · Answer 11 · 2016-02-02T04:58:14.183

1

You could do this by Regular Expression as follows you could use this code

EditText et = (EditText) findViewById(R.id.editText);
String NumberPattern = "[0-9]+";
String Number = et.getText().toString();

if (Number.matches(NumberPattern) && s.length() > 0)
{ 
    //code for number
}
else
{
    //code for incorrect number pattern
}

edited Feb 02 '16 at 04:58

answered Jan 28 '16 at 12:50

Jeffy

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et is of editText. + sign is for the occurances of digits from 0 to 9 – Jeffy Jan 28 '16 at 13:42

score 0 · Answer 12 · edited Jan 28 '16 at 12:57

I have coded a sample program that checks if a string contains a number in it! I guess it will serve for this purpose as well.

public class test {
    public static void main(String[] args) {
        String c;
        boolean b;

        System.out.println("Enter the value");
        Scanner s = new Scanner(System.in);
        c = s.next();
        b = containsNumber(c);
        try {
            if (b == true) {
                throw new CharacterFormatException();
            } else {
                System.out.println("Valid String \t" + c);
            }
        } catch (CharacterFormatException ex) {
            System.out.println("Exception Raised-Contains Number");

        }
    }

    static boolean containsNumber(String c) {
        char[] ch = new char[10];
        ch = c.toCharArray();

        for (int i = 0; i < ch.length; i++) {
            if ((ch[i] >= 48) && (ch[i] <= 57)) {
                return true;
            }
        }
        return false;
    }
}

CharacterFormatException is a user defined Exception. Suggest me if any changes can be made.

How to check if a character in a string is a digit or letter

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