Java login system method

Question

Have almost got this login system in Java to work, but am having trouble with this method:

public void Register() {
    Scanner sc = new Scanner(System.in);

    System.out.print("Register? (Y/ N)\n");
    String N = sc.nextLine();

    if ("N".equals(N)) {
        Login();
    } else {
        String Y = sc.nextLine();
        if ("Y".equals(Y)) {
        System.out.print("Email address: ");
        String string = sc.nextLine();
        System.out.print("Password: ");

        String string2 = sc.nextLine();

        System.out.print("\n\n");
        new Products().search();
        }
    }

}

Entering "N" works perfectly in the if section, but "Y" needs to be entered twice before the else section works (I understand why it doesn't work).

Its something simple I know, but any clue as to how to get it to work?

Appreciate any help...

Please please please: never name variables N or Y. Use name the mean something to human readers. Such as: inputFromUser. Just N N N. — GhostCat
– GhostCat, Commented Jul 9, 2018 at 11:51
As the best coding practices, please rename your variables to more intuitive names. Like String userInput = sc.nextLine(); String emailId = sc.nextLine(); and so on. — Deepak Agrawal
– Deepak Agrawal, Commented Jul 9, 2018 at 11:54

khelwood · Accepted Answer · 2018-07-09 11:49:55Z

5

Here

String Y = sc.nextLine();

you are reading in another line of input. You want to compare the same line of input you already read, which you stored in a variable called N. It will be clearer if you give it a better name.

String line = sc.nextLine();

if ("N".equals(line)) {
    Login();
} else if ("Y".equals(line)) {
    System.out.print("Email address: ");
    ...
}

answered Jul 9, 2018 at 11:49

khelwood

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1 Comment

user1924813 Over a year ago

Thanks this was what I was looking for

duke707 · Accepted Answer · 2018-07-09 12:14:15Z

3

This String N = sc.nextLine(); Again This

String Y = sc.nextLine();

No need to use the input method twice

Change the variable name to something meaningful

Try it this way

It will surely work..

public void Register() {
Scanner sc = new Scanner(System.in);

System.out.print("Register? (Y/ N)\n");
String input = sc.nextLine();

if ("N".equals(input)) {
    Login();
} else {
    // removed 'String Y = sc.nextLine();'
    if ("Y".equals(input)) {
    System.out.print("Email address: ");
    String string = sc.nextLine();
    System.out.print("Password: ");

    String string2 = sc.nextLine();

    System.out.print("\n\n");
    new Products().search();
    }
}

answered Jul 9, 2018 at 12:14

duke707

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Comments

Murat Karagöz · Accepted Answer · 2018-07-09 11:50:02Z

1

You don't need the second nextLine. Keep using the String N (maybe rename it to input) and keep checking that for Y.

answered Jul 9, 2018 at 11:50

Murat Karagöz

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Comments

azro · Accepted Answer · 2018-07-09 11:56:55Z

1

If you type Y it will be saved in N variable, it will go in the else section and here you're asking again the user, you don't need to

You have to take the input and test it, without knowing its N or Y

String choice = sc.nextLine();

if ("N".equals(choice)) {
    Login();
} else if("Y".equals(choice)){      
    System.out.print("Email address: ");
    String string = sc.nextLine();
    ...       
}else{
     System.out.println("Wrong choice");
}

Also

give more significant names to your variable, no string, string2 but email, pwd
methods name have to start in lowerCaser : Login() >> login()

edited Jul 9, 2018 at 11:56

answered Jul 9, 2018 at 11:51

azro

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