What is the best way to tell if a character is a letter or number in Java without using regexes?

Question

What is the best and/or easiest way to recognize if a string.charAt(index) is an A-z letter or a number in Java without using regular expressions? Thanks.

Answer 1

Character.isDigit(string.charAt(index)) (JavaDoc) will return true if it's a digit
Character.isLetter(string.charAt(index)) (JavaDoc) will return true if it's a letter

Answer 2

I'm looking for a function that checks only if it's one of the Latin letters or a decimal number. Since char c = 255, which in printable version is ├ and considered as a letter by Character.isLetter(c). This function I think is what most developers are looking for:

private static boolean isLetterOrDigit(char c) {
    return (c >= 'a' && c <= 'z') ||
           (c >= 'A' && c <= 'Z') ||
           (c >= '0' && c <= '9');
}

Answer 3

As the answers indicate (if you examine them carefully!), your question is ambiguous. What do you mean by "an A-z letter" or a digit?

• If you want to know if a character is a Unicode letter or digit, then use the Character.isLetter and Character.isDigit methods.

• If you want to know if a character is an ASCII letter or digit, then the best thing to do is to test by comparing with the character ranges 'a' to 'z', 'A' to 'Z' and '0' to '9'.

Note that all ASCII letters / digits are Unicode letters / digits ... but there are many Unicode letters / digits characters that are not ASCII. For example, accented letters, cyrillic, sanskrit, ...

---

The general solution is to do this:

Character.UnicodeBlock block = Character.UnicodeBlock.of(someCodePoint);

and then test to see if the block is one of the ones that you are interested in. In some cases you will need to test for multiple blocks. For example, there are (at least) 4 code blocks for Cyrillic characters and 7 for Latin. The Character.UnicodeBlock class defines static constants for well-known blocks; see the javadocs.

Note that any code point will be in at most one block.

Answer 4

Java Character class has an isLetterOrDigit method since version 1.0.2

Answer 5

I don't know about best, but this seems pretty simple to me:

Character.isDigit(str.charAt(index))
Character.isLetter(str.charAt(index))

Answer 6

// check if ch is a letter
if ((ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z'))
    // ...

// check if ch is a digit
if (ch >= '0' && ch <= '9')
    // ...

// check if ch is a whitespace
if ((ch == ' ') || (ch =='\n') || (ch == '\t'))
    // ...

Source: https://docs.oracle.com/javase/tutorial/i18n/text/charintro.html

Answer 7

Use the below code

Character.isLetterOrDigit(string.charAt(index))

Answer 8

Compare its value. It should be between the value of 'a' and 'z', 'A' and 'Z', '0' and '9'

Answer 9

 import java.util.Scanner;
 public class v{
 public static void main(String args[]){
 Scanner in=new Scanner(System.in);
    String str;
    int l;
    int flag=0;
    System.out.println("Enter the String:");
    str=in.nextLine();
    str=str.toLowerCase();
    str=str.replaceAll("\\s","");
    char[] ch=str.toCharArray();
    l=str.length();
    for(int i=0;i<l;i++){
        if ((ch[i] >= 'a' && ch[i]<= 'z') || (ch[i] >= 'A' && ch[i] <= 'Z')){
        flag=0;
        }
        else

        flag++;
        break;
        } 
if(flag==0)
    System.out.println("Onlt char");


}
}