CODEWITHSUNDEEP
javaif-statementjava.util.scanneruser-input
justsomerandomguy
justsomerandomguy

Reputation: 33

Java Scanner input with if else statement

Hi I'm new to java and trying to make a Quiz to practice. I wanna make a Question where the user has to combine words from to categories to pairs. Like A1 B4 C3 D2. What I did now is using an if else statement to check if the input is the correct answer, but it only works with 1A. For the others I can do 6 inputs, which is not what I want, and even if there's a correct one I don't get a point.

public class HelloWorld {

    public static void main(String[] args) {

        Scanner walther = new Scanner(System.in);

        String cro = "1A";
        String dan = "2C";
        String fin = "4D";
        String dut = "3F";
        String fre = "5B";
        String ger = "6E";
        int x = 0;


        if (cro.equalsIgnoreCase(walther.nextLine())){
            ++x;
            walther.close();
        }
        else if (dan.equalsIgnoreCase(walther.nextLine())){
            ++x;
            walther.close();
        }
        else if (fin.equalsIgnoreCase(walther.nextLine())){
            ++x;
            walther.close();
        }
        else if (dut.equalsIgnoreCase(walther.nextLine())){
            ++x;
            walther.close();
        }
        else if (fre.equalsIgnoreCase(walther.nextLine())){
            ++x;
            walther.close();
        }
        else if (ger.equalsIgnoreCase(walther.nextLine())){
            ++x;
            walther.close();
        }
        else {
            walther.close();
        }

    System.out.println(x + " Point!");
    }
}

Upvotes: 2

Views: 20444

Answers (2)

Shenshow Shodin
Shenshow Shodin

Reputation: 1

People the fact remains that the above provided answer is confusing especially to those who are relatively new to the java community. Hence the requirement is for an easier and simpler answer. Now, Java understands Strings only by the following code :

Scanner sc=new Scanner(System.in);
String a=sc.next();

if (a.equals("xyzzy"))
{
  System.out.println("yes");

Upvotes: -1

Mureinik
Mureinik

Reputation: 312344

Calling nextLine() consumes a line from the scanner. You do this on the first if, so the subsequent else if branches are, in fact, comparing the following lines (or null, if you don't have any additional input). Instead, you should consume the line only once, save it to a local variable and use that in your comparisons:

String input = walther.nextLine();
if (cro.equlasIgnoreCase(input)) { // etc...

Having said that, using and if-else structure isn't the neatest solution. You can save a lot of code bloat by using a case insensitive TreeSet:

TreeSet<String> set = new TreeSet<>(String.CASE_INSENSITIVE_ORDER);
set.addAll(Arrays.asList("1A", "2C", "4D", "3F", "5B", "6E"));
String input = walther.nextLine();
if (set.contains(input)) {
   ++x;
   walther.close();
}

Upvotes: 3

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