使用Character.isLetter和Character.isDigit忽略数字，空格和读取字符串输入

Question

I have been stuck for quite some time. I would need to ignore digits entered, blank spaces and special characters like $^%##&! and just read the other letters az, using Character.isDigit & Character.isLetter.. I have tried using the both methods, it didn't work out for me.. Please advice..

The error:

The normal output (without spaces and digits) :

The expected output should be 438-5626 even when I entered 123$@GetLoan.. They should ignore the first few characters '123$@' and read only GetLoan..

Full Question: Write a program that prompts the user to enter a telephone number expressed in letters and outputs the corresponding telephone number in digits. If the user enters more than seven letters, then process only the first seven letters. Also output the – (hyphen) after the third digit. Allow the user to use both uppercase and lowercase letters as well as spaces between words.

public class Question3 {

public static void main(String[] args) {

    String letters;
    char phoneDigit;

    Scanner kb = new Scanner(System.in);

    System.out.println("Enter letters : ");
    letters = kb.next();

    for (int i = 0; i < 7; i++) {

        phoneDigit = letters.charAt(i);

        if (Character.isLetter(phoneDigit) == true) {

            if (i == 3) {
                System.out.println("-");
            } //If 

            if (phoneDigit >= 'A' && phoneDigit <= 'C'
                    || phoneDigit >= 'a' && phoneDigit <= 'c') {

                System.out.println("2");

            } else if (phoneDigit >= 'D' && phoneDigit <= 'F'
                    || phoneDigit >= 'd' && phoneDigit <= 'f') {

                System.out.println("3");

            } else if (phoneDigit >= 'G' && phoneDigit <= 'I'
                    || phoneDigit >= 'g' && phoneDigit <= 'i') {

                System.out.println("4");

            } else if (phoneDigit >= 'J' && phoneDigit <= 'L'
                    || phoneDigit >= 'j' && phoneDigit <= 'l') {

                System.out.println("5");

            } else if (phoneDigit >= 'M' && phoneDigit <= 'O'
                    || phoneDigit >= 'm' && phoneDigit <= 'o') {

                System.out.println("6");

            } else if (phoneDigit >= 'P' && phoneDigit <= 'S'
                    || phoneDigit >= 'p' && phoneDigit <= 's') {

                System.out.println("7");

            } else if (phoneDigit >= 'T' && phoneDigit <= 'V'
                    || phoneDigit >= 't' && phoneDigit <= 'v') {

                System.out.println("8");

            } else if (phoneDigit >= 'W' && phoneDigit <= 'Z'
                    || phoneDigit >= 'W' && phoneDigit <= 'z') {

                System.out.println("9");
            } // If
        } // If
    } // For loop

} //PSVM

Answer 1

Below fragment is probably the problem:

for (int i = 0; i < 7; i++) {
    phoneDigit = letters.charAt(i);
    if (Character.isLetter(phoneDigit) == true) {

You are taking first 7 characters, and printing only the ones that are letters. So for input string 123getloan you will iterate over 123getl , and then isLetter will reject 123 , so your program later will deal only with getl .

To iterate over only 7 letters you would need to change it to increment i only if given character is a letter eg by doing the below:

int i = 0;
for (char phoneDigit : letters.toCharArray()) {
    if (Character.isLetter(phoneDigit)) {
        i++;

        // other ifs here

        if (i == 3) {
            System.out.println("-");
        }
    }
    if (i >= 7) {
        break;
    }
}

EDIT: Fixed problem mentioned by @Andreas