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  1. Hi, I attached my solution about yours problems, I comment diferents parts about code. There are two functions where a signal if is a numérica data or category of age. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

    import java.util.Scanner;

public class main {

public static void main(String[] args) {
    System.out.print("Type your age: ");

    Scanner sc = new Scanner(System.in);
//Save var String
    String ageStr = sc.nextLine();

    String notify="Not an age.";
    //Evaluate the age if we typify a number.
        if (isNumeric(ageStr)==true) {
            int age = Integer.parseInt(ageStr);
            notify = myAge(age);
        }

    //Notification about your Age
        System.out.println(notify);
}
//Method about Age
public static String myAge(int age) {

    if (age >= 18) {
        return "Over 18 or equals 18";
    }
    else if (age == 0) {
        return "0 age?";
    }
    else if (age <= 18) {
        return "Under 18";
    }
    return null;
}
//Method is numeric?
public static boolean isNumeric(final String str) {

    if (str == null || str.length() == 0) {
        return false;
    }

    for (char c : str.toCharArray()) {
        if (!Character.isDigit(c)) {
            return false;
        }
    }

    return true;

}

    }

I hope that can I help you

Regards

Chmoro
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Bruno
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  • "It doesn't work" is not a meaningful problem statement. Please [edit] your question to include the details of *how* your program doesn't work. – azurefrog Sep 24 '19 at 20:13
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    Your code snippet doesn't show any attempt to use `String age2 = Integer.toString(age)`, which makes it difficult for us to tell what you're doing wrong. – Dawood ibn Kareem Sep 24 '19 at 20:15
  • Use camel case for methods in Java. – arcadeblast77 Sep 24 '19 at 20:21
  • Your problem is not that you can't convert an int to String, your problem is that you don't properly check the input prior reading it as int. I'll link you a question which tells you how to do that. – Tom Sep 24 '19 at 20:22
  • If i just return "Not an age" this happens: Type your age: dsadsa Exception in thread "main" java.util.InputMismatchException at java.base/java.util.Scanner.throwFor(Scanner.java:939) at java.base/java.util.Scanner.next(Scanner.java:1594) at java.base/java.util.Scanner.nextInt(Scanner.java:2258) at java.base/java.util.Scanner.nextInt(Scanner.java:2212) at Prova.Age.AgeFunction(Age.java:14) at Prova.Test.main(Test.java:8) Process finished with exit code 1 – Bruno Sep 24 '19 at 20:23
  • `nextInt` throws exceptions. You can handle them with `try`/`catch`. The docs tell you which exceptions `nextInt` throws and why. https://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html#nextInt() – arcadeblast77 Sep 24 '19 at 20:26

1 Answers1

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As you said, you'll currently get an InputMismatchException. You could simply catch this and return your desired value.

Doc for Scanner.nextInt() linked by arcadeblast77

public class Age {

    public String AgeFunction() {

        Scanner sc = new Scanner(System.in);

// User type your age
        try {
            System.out.print("Type your age: ");
            int age = sc.nextInt();

            if (age >= 18) {
                return "Over 18 or equals 18";
            }

            else if (age == 0) {
                return "0 age?";
            }

            else if (age <= 18) {
                return "Under 18";
            }

    // If what was typped be different int, return string


        } catch (InputMismatchException exception) {
            return "Not an age.";
        }

    }

}
DerMolly
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