学习JOptionPane有麻烦吗？

Question

Instead of System.out.println for everything, I want to use JOptionPane. I don't know how to describe how my program is skipping code. If you could please take a look at this and help it go in order that would be highly appreciated.

//Step 1: Import Java APIs
 import java.util.InputMismatchException;
 import java.util.Scanner;
 import javax.swing.*;



//Step 2: Name File and Class
 public class GolfScores {




//Step 3: Declare All Variables  

 public static void main(String[] args) {
  int hole9 = 0;
  int hole18 = 0;
  String holeChoice;

  Scanner input = new Scanner(System.in);


//Step 4: The program accepts INPUT from the user

  for (;;) {
   try {
 holeChoice =   JOptionPane.showInputDialog("For hole 9, please enter 9. If you are on hole 18, please enter 18 ");

     if (holeChoice.equals(9)) {
        System.out.println("The par for this hole is 3. Please enter if you this hole took you: "
                    + "1, 2, 3, 4, 5, 6+ shots");            
        hole9 = input.nextInt();

    } else if (holeChoice.equals(18)) {
        System.out.println("The par for this hole is 5. Please enter if you this hole took you: "
                    + "1, 2, 3, 4, 5, 6, 7 or 8+ shots");
        hole18 = input.nextInt();

    } else if (holeChoice.equals(18)) {
      System.out.println("Please enter a valid number");
      continue;
    }
    break;
} catch (InputMismatchException inputMismatchException) {
    System.err.printf("\nException: %s\n", inputMismatchException);
    System.out.println("Please enter a valid number.");
    input.next();
}
}


//Step 5 & 6: The user input is PROCESSED by Java and uses math to  calcualte an answer to output. The user's score is then output for them to see.
 if (hole18 == 1) {
   System.out.println("Your score for this hole was a hole in one!");

 } else if (hole18 == 2) {
   System.out.println("Your score for this hole was albetross.");

 } else if (hole18 == 3) {
   System.out.println("Your score for this hole was eagle.");

 } else if (hole18 == 4) {
   System.out.println("Your score for this hole was birdie.");

 } else if (hole18 == 5) {
   System.out.println("Your score for this hole was par.");

 } else if (hole18 == 6) {
   System.out.println("Your score for this hole was boogie.");

 } else if (hole18 == 7) {
   System.out.println("Your score for this hole was double boogie.");

 } else if (hole18 >= 8) {
   System.out.println("You need some more practice at golf.");

 } else if (hole18 <= 0) {
   System.out.println("Please input a valid number.");
   input.next();

 }


   if (hole9 == 1) {
   System.out.println("Your score for this hole was a hole in one!");

 } else if (hole9 == 2) {
   System.out.println("Your score for this hole was birdie.");

 } else if (hole9 == 3) {
   System.out.println("Your score for this hole was par.");

 } else if (hole9 == 4) {
   System.out.println("Your score for this hole was boogie.");

 } else if (hole9 == 5) {
   System.out.println("Your score for this hole was double boogie.");

 }  else if (hole9 >= 6) {
   System.out.println("Your need some more practice at golf.");

  }

 try {
  Thread.sleep(3000);                
}     catch(InterruptedException ex) {
      Thread.currentThread().interrupt();
}

    System.out.println("Thank you for playing Java Golf. Please come      again");

   }

}

Answer 1

The returned value from JOptionPane.showInputDialog(..) are String so use:

 if (holeChoice.equals("9")) { ... }

Or either parse the String to Integer using:

Integer.parseInt(holeChoice);

Answer 2

JOptionPane.showInputDialog returns a String , you're comparing the result as if it was an Integer .

holeChoice =   JOptionPane.showInputDialog("For hole 9, please enter 9. If you are on hole 18, please enter 18 ");
//...
if (holeChoice.equals(9))  // this will always be false

If the user enters 9 , comparing it to the int 9 will not work, you need to add quotes to the values you check.

Ex:

if (holeChoice.equals("9"))

Likewise, you could also try to parse the input and turn it to an int , but you might open up the door for invalid user input (to which you'd need to check)

try {
   Integer choice = Integer.valueOf(holeChoice);
   //...
   if (choice.equals(9)) { // works now...
}
catch (NumberFormatException ex) {
   // user didn't enter a number!
}