How would I be able to determine if a string contains only one digit?

Abby :

I would need to find if the user input for password contains only 1 digit and I am having trouble what the if statement would be.

I currently only have:

import java.util.Scanner;

public static boolean onlyOneDigit (String password) {
    if ( ........ ) {
        System.out.print("Password requirement not met");
        return true;
    } else {
        return false;
    }
}

public static void main(String[] args) {
    do {
        System.out.print("Enter a password");
        String password = input.next();
    } while (onlyOneDigit(password)); 

    {
        System.out.print("Password updated");
    }
}

WJS :

Try this. Single digit in password vs no digit or more than one in password.

        for (String pw : new String[]{"with1digit", "withtwo22digits","withNodigits",
                "with2scatt2reddi3gits"}) {
            boolean m = pw.matches("[^\\d]*\\d[^\\d]*");
            System.out.println(m + " : " + pw);
        }

Prints the following:

true : with1digit
false : withtwo22digits
false : withNodigits
false : with2scatt2reddi3gits

Example usage.

     if (onlyOneDigit(password)) {
         // it's good
     } else {
         // warn user
     }

     public static boolean onlyOneDigit(String pw) {
         return pw.matches("[^\\d]*\\d[^\\d]*");
     }

And her is another way

        public static boolean onlyOneDigit(String pw) {
            // get rid of the first digit
            String save =pw.replaceFirst("\\d","");
            // replace the next digit with a # 
            save = save.replaceFirst("\\d", "#");
            // if save contains # it had more than 1 digit
            // or if save equals the original password
            // then there were no digits.
            return !(save.equals(pw) || save.contains("#"));
         }

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