使用开关和扫描仪级Java

Question

我在使用扫描仪类的从键盘 我试图从主方法

StudentBO country = new StudentBO(); 
Scanner sc = new Scanner(System.in); 
int selection; 
System.out.println("Menu : "); 
System.out.println("Type any number between 1 and 6"); 
System.out.println("1)Create a new student"); 
System.out.println("2)details specific student"); 
System.out.println("3)details of all students"); 
System.out.println("4)details of the student age details"); 
System.out.println("5)details of the student personal info"); 
System.out.println("6)Exit"); 
selection = sc.nextInt(); 
switch (selection) 
{ 
    case 1 : 
     System.out.println("Enter studentName "); 
     sc = new Scanner(System.in); 
     String studentName= sc.nextLine(); 
     Student studentInfo = student.createStudent(studentName); 
     System.out.println("Do you want to continue? Type Yes/No"); 
     Scanner sc1 = new Scanner(System.in); 

    case 2 : 
     break; 
    case 3 : 
     break; 
    case 4 :  
     break; 
    case 5 : 
     break; 
    case 6 : 
     break; 
} 

当我执行下面的逻辑选择一些菜单有一些问题

使用用户选择的选择选择选项1，创建学生，然后我必须输入是或否，如果用户选择是，则再次从选项菜单（1,2,3,4,5或6）中提示用户，如果用户选择否，然后休息。

问题我面临的

1. 如何退一万例的执行之后要选择的选项菜单。我想回到提示用户选择菜单后的情况1

2. 如何从一种情​​况切换到另一种情况。在情况1中，基于某些条件，我想直接调用情况2。那可能吗？

Answer 1

您可以在读取第二个输入时使用while循环和布尔值退出。

StudentBO country = new StudentBO(); 
Scanner sc = new Scanner(System.in); 
int selection; 
boolean exit = false; 
while (!exit) 
{ 
System.out.println("Menu : "); 
System.out.println("Type any number between 1 and 6"); 
System.out.println("1)Create a new student"); 
System.out.println("2)details specific student"); 
System.out.println("3)details of all students"); 
System.out.println("4)details of the student age details"); 
System.out.println("5)details of the student personal info"); 
System.out.println("6)Exit"); 
selection = sc.nextInt(); 
switch (selection) 
     { 
     case 1 : 
      System.out.println("Enter studentName "); 
      sc = new Scanner(System.in); 
      String studentName= sc.nextLine(); 
      Student studentInfo = student.createStudent(studentName); 
      System.out.println("Do you want to continue? Type Yes/No"); 
      Scanner sc1 = new Scanner(System.in); 
      if (sc1.next().equalsIgnoreCase("no")) 
        exit = true; 
       break; 

      case 2 : 

       break; 
      case 3 : 

       break; 
      case 4 : 

       break; 
      case 5 : 
       break; 
      case 6 : 
       break; 
      } 
}