Checking for Alphabets in a String in java

Question

I have a string \"BC+D*E-\". I want to check each character of the string whether it is an alphabet or not. I tried using isLetter() but it does consider even the = , * and

Answer 1

Use the Wrapper class Character#isLetter(char) method

Character.isLetter(char c);

and check each letter one at a time.

Answer 2

String stringEx =  "BC+D*E-";
for (char string : stringEx.toCharArray()) {
    if ( ( (char)string > 64 ) && ((char)string < 91) )
        System.out.println("It is character");
    if ( ( (char)string > 96 ) && ((char)string < 123) )
    System.out.println("It is character");
}

You can use this code. This may also helpful to you.

Answer 3

This snippet may help you.

 String startingfrom = "BC+D*E-".toUpperCase();
        char chararray[] = startingfrom.toCharArray();
        for(int i = 0; i < chararray.length; i++) {
                            int value = (int)chararray[i];
                            if((value >= 65 && value <= 90) || (value >= 97 && value <= 122))
                                System.out.println(chararray[i]+ " is an alphabate");
                            else 
                                System.out.println(chararray[i]+ " is not an alphabate");
        }

Answer 4

You can use ASCII value of English letters. e.g.

for (char c : "BC+D*E-".toCharArray())
{
  int value = (int) c;
  if ((value >= 65 && value <= 90) || (value >= 97 && value <= 122))
  {
    System.out.println("letter");
  }
}

You can find ASCII table here: http://www.asciitable.com/

Answer 5

Try

    String s = "BC+D*E-=";

    for (int i = 0; i < s.length(); i++) {
        char charAt2 = s.charAt(i);
        if (Character.isLetter(charAt2)) {
            System.out.println(charAt2 + "is a alphabet");
        }
    }

Answer 6

Break the String into an array and iterate over it.

String word = "BC+D*E-"
for (char c : word.toCharArray()) {
    if(!(Character.isLetter(c)) {
        System.out.println("Not a character!");
        break;
    }
}