Switch語句未退出循環。 （Java）

Question

我有一個簡單的菜單系統，用戶點擊8即可離開。 但是由於某種原因，當我在測試過程中達到8時，它就像沒有任何反應一樣，回到了循環的頂部。

package potluck;
import java.util.*;

import potluck.*;

public class Controller {
private Scanner input;


private final static int USER_LOGIN = 0;
private final static int CREATE_MEMBER = 1;
private final static int CREATE_ADMIN = 2;
private final static int CREATE_RECIPE = 3;
private final static int COMMENT = 4;
private final static int DELETE_RECIPE = 5;
private final static int EXIT = 8;

public Controller(){
    input = new Scanner(System.in);
    startUp();//no better name to be thought of
}

public void startUp() {
    // TODO Auto-generated method stub
    int choice;
    do {
        this.displayMenu();
        choice = input.nextInt();
        input.nextLine();// clears carriage return

        //depending on choice takes to a different menu
        switch (choice) {

        case CREATE_MEMBER: 
            Member member = new Member();
            break;
//     case CREATE_ADMIN: 
//      member.addAdmin();
//              break;
        case CREATE_RECIPE:
            Recipe.addRecipe();
            break;
        case COMMENT:
            Recipe.addComment();
            break;
        case DELETE_RECIPE:
            Recipe.deleteRecipe();
            break;
        case EXIT:
            System.out.println("Thanks for using our software");
            break;
        default:
            System.out.println("Error, Invalid selection.");
        }
    } while (choice != 8); //choice 8 exits
}
private void displayMenu() {
    System.out.println("1 Create Member");
    System.out.println("2 Create Admin Member");
    System.out.println("3 Create Recipe");
    System.out.println("4 Leave Comment");
    System.out.println("5 Delete Recipe");
    System.out.println("8 Exit");
    System.out.println("Please enter menu option, to exit enter 8");
    }
}

在測試中，它聲稱選擇為8，這應該會打破一會兒……但不會……

更新：復制代碼時，我被告知不要使用某些工作。 我在選擇項8下有system.exit，但被告知那是錯誤的代碼

Answer 1

不好意思 我是個白痴。 在啟動它的主要位置，我實例化了一個新的控制器（您正在查看的類），該控制器在構造函數中啟動菜單。 然后我調用了startup（）方法。 這意味着它運行良好，因為我很笨，它只運行了兩次。

public static void main(String[] args) {
    // TODO Auto-generated method stub
    Controller cntlr = new Controller();    
    cntlr.startUp();
}


public Controller(){
    input = new Scanner(System.in);
    startUp();//no better name to be thought of
}

public void startUp() {
    // TODO Auto-generated method stub
    int choice;
    do {
        this.displayMenu();
        choice = input.nextInt();
        input.nextLine();// clears carriage return

        //depending on choice takes to a different menu
        switch (choice) {

        case CREATE_MEMBER: 
            Member member = new Member();
            break;
        case CREATE_ADMIN: 
            member.addAdmin();
            break;
        case CREATE_RECIPE:
            Recipe.addRecipe();
            break;
        case COMMENT:
            Recipe.addComment();
            break;
        case DELETE_RECIPE:
            Recipe.deleteRecipe();
            break;
        case EXIT:
            System.out.println("Thanks for using our software");

            break;
        default:
            System.out.println("Error, Invalid selection.");
        }
    } while (choice != 8); //choice 8 exits
}
private void displayMenu() {
    System.out.println("1 Create Member");
    System.out.println("2 Create Admin Member");
    System.out.println("3 Create Recipe");
    System.out.println("4 Leave Comment");
    System.out.println("5 Delete Recipe");
    System.out.println("8 Exit");
    System.out.println("Please enter menu option, to exit enter 8");
    }
}

TL; DR。 我很傻，構造函數是我應該更多注意的事情。