java - 菜单程序中的可变错误

Question

我想创建一个菜单程序，具有不同的选项:

1)Enter Name
2)Display Name 
3)Change Name
4)Quit

但由于某种原因，我无法获取用于在 Q1 中存储名称的变量，以便在 Q2 中使用!

代码如下:(错误发生在情况2中，变量full_name带有红色下划线并且不起作用)

package Testing;
import java.util.Scanner ;
public class Menu {

public static void main(String[] args) {
    // TODO Auto-generated method stub
    Scanner in = new Scanner(System.in);
    // print menu
    for (int i = 1; i <= 5; i++)
          System.out.println(i + ". Menu item #" + i);
    System.out.println("0. Quit");
    // handle user commands
    boolean quit = false;
    int menuItem;
    do {
          System.out.print("Choose menu item: ");
          menuItem = in.nextInt();
          switch (menuItem) {
          case 1:
                System.out.println("You've chosen item #1");
                {
                    Scanner user_input = new Scanner(System.in);

                    String first_name;
                    System.out.print("Enter your first name: ");
                    first_name = user_input.next( );

                    String family_name;
                    System.out.print("Enter your family name: ");
                    family_name = user_input.next( );

                    String full_name;
                    full_name = first_name + " " + family_name;

                    System.out.println("You are " + full_name);
                }

                break;
          case 2:
                System.out.println("You've chosen item #2");
                System.out.println("You are " + full_name);
                break;
          case 3:
                System.out.println("You've chosen item #3");
                // do something...
                break;
          case 4:
                System.out.println("You've chosen item #4");
                // do something...
                break;
          case 5:
                System.out.println("You've chosen item #5");
                // do something...
                break;
          case 0:
                quit = true;
                break;
          default:
                System.out.println("Invalid choice.");
          }
    } while (!quit);
    System.out.println("Bye-bye!");
 }
}

Answer 1

您应该将 full_name 放在循环之外，因为当前它位于本地范围内，无法从其他 case 语句访问。