Why cant I input info after if statement? No errors show

Question

How can I get the program to ask for input using the scanner after the if statement?

import java.util.Scanner;

public class App1 {

    public static void main(String[] Args) {

        Scanner keyboard = new Scanner(System.in);

        String gender;
        String fName;
        String lName;
        int age;
        String female = "F";
        String male = "M";
        String a = "";
        String gettingMarried = "y";
        String notGettingMarried = "n";

        System.out.println("What is your gender (M or F)");
        gender = keyboard.nextLine();
        System.out.println("What is your first name?");
        fName = keyboard.nextLine();
        System.out.println("What is your last name?");
        lName = keyboard.nextLine();
        System.out.println("Age:");
        age = keyboard.nextInt();
        System.out.println();

        if (gender.equals(female) && age >= 20) {
            System.out.println("Are you married " + fName + " (y or n?)");
            a = keyboard.nextLine();

        } else if (gender.equals(male) && age >= 20) {
            System.out.println("Are you married" + fName + "(y or n?)");
            a = keyboard.nextLine();
        }

    }
}

1. It seems like you can simplify your if statement to age >= 20 but I might be missing something. 2. But more to the point of your question, is it printing the line: "Are you married .... "? — Nikhil Agarwal, Aug 25 '15 at 00:28
What would only happen if neither case is being satisfied. For example if gender was lowercase 'm'/'f' or age was less than 20. — Gary, Aug 25 '15 at 00:28
What is your gender (M or F) F What is your first name? Jerry What is your last name? Woods Age: 24 Are you married Jerry (y or n?) — Jerry Woods, Aug 25 '15 at 01:24

score 1 · Accepted Answer · edited May 23 '17 at 12:22

1

It is because you use a nextInt() that only takes the int value and not the full line. When you try again to read with keyboard.nextLine() what you are catching it's the rest of the sentence of the line in which you put the int.

Change the System.out.println(); before your if statement to keyboard.nextLine();

Edit: If you want a more extended explanation about why you have to put keyboard.nextLine(); after your keyboard.nextInt() you can consult my answer in another question with the same problem: Why isn't the scanner input working?

I expect it will be helpful for you!

edited May 23 '17 at 12:22

Community

1
1

answered Aug 25 '15 at 00:31

Francisco Romero

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But the age is a the age which is a int. I get an error when I change it. – Jerry Woods Aug 25 '15 at 01:27
Oh no, you don't have to change it. You have to let, as you have, `age = keyboard.nextInt();`. You just have to change the line `System.out.println();` to `keyboard.nextLine()` without storing the value. – Francisco Romero Aug 25 '15 at 01:30
Thanks worked well, I'm just trying to figure out what "keyboard.nextLine();" does. – Jerry Woods Aug 25 '15 at 01:36
@JerryWoods Look my other answer (in the link of the question that I put above). If it doesn't clearify your doubts just ask to me ;) – Francisco Romero Aug 25 '15 at 01:37

Why cant I input info after if statement? No errors show

1 Answers1