Question

出于某种原因，在我添加名为“Oliver”的宠物后，主菜单会打印出两次“无效选择”行。我只需要另外一双眼睛来看它，因为我已经连续几个小时看着它，并且修复了一些小错误但无济于事。

运行时的代码如下所示：

     /*Welcome to the pet store.Type the letter to make your selection
      A.  List the pets in the store.
      B.  Age up the pets
      C.  Add a new pet
      D.  Adopt a pet
      E.  Quit
      C
      Please type in a name
      Oliver
      Please type in an age
      22
      Oliver has just been added to the store!
      Welcome to the pet store.Type the letter to make your selection
      A.  List the pets in the store.
      B.  Age up the pets
      C.  Add a new pet
      D.  Adopt a pet
      E.  Quit
      Invalid choice
       Welcome to the pet store.Type the letter to make your selection
       A.  List the pets in the store.
       B.  Age up the pets
       C.  Add a new pet
       D.  Adopt a pet
       E.  Quit*/

这是我的主要类代码：

    private static void mainmenu(){
    System.out.println("Welcome to the pet store.Type the letter to make 
    your selection");
    System.out.println("A."+"  " + "List the pets in the store.");
    System.out.println("B."+"  " + "Age up the pets");
    System.out.println("C."+"  " + "Add a new pet");
    System.out.println("D."+"  " + "Adopt a pet");
    System.out.println("E."+"  " + "Quit");

    MainPets.Getuserinput();

}

public static String Getuserinput(){

    userinput=scan.nextLine();

    return userinput; 

}

   public static void main (String [] args){
    int pet3age;
    String pet3name;
    Pet Pet1=new Pet("Fido",3); 
    Pet Pet2=new Pet("Furball",1);
    Pet Pet3=null;
    int userinputint;

    MainPets.mainmenu();


     while(userinput.equals("A")||userinput.equals("B")||userinput.equals("C")||userinput.equals("D")||userinput.equals("E")){

         switch(userinput){
         case "C":

            if (Pet3!=null&&userinput.equals("C")){
                System.out.println("Sorry the store is full");
            }

            if(Pet3==null){ 
                System.out.println("Please type in a name");
                pet3name=scan.nextLine();
                System.out.println("Please type in an age");
                pet3age=scan.nextInt();
                Pet3=new Pet(pet3name,pet3age);
                System.out.println(pet3name + " has just been added to the store!");
            }
            MainPets.mainmenu();
            break;
            }
            }
        while(!userinput.equals("A")||!userinput.equals("B")||!userinput.equals("C")||!userinput.equals("D")||!userinput.equals("E")){
      System.out.println("Invalid choice");
      MainPets.mainmenu();
   }

这是包含所有方法的类：

public class Pet {
String Name; 
String AdoptionStatus; 
int Age;

public Pet() {}

public Pet(String Name, int Age) {
    this.Name = Name;
    this.Age = Age;
}

public void SetName(String namesetup) {
    Name = namesetup;
}

public String GetName() {
    return Name;
}

public int GetAge() {
    return Age;
}

public int ageincrease() {
    return Age++;
}

public String Getadoptionstatus() {
    return AdoptionStatus;
}

 public void Setadoptionstatustonotadopted(int petnumber) {
    AdoptionStatus="not adopted";
}

public void Setadoptionstatustoadopted(int petnumber){
    AdoptionStatus="adopted";
}

}

Answer 1

您似乎正在尝试尽可能多地使用static来实践它的功能吗？

无论如何，请参阅下面的一个最小的例子，你可以建立一个最小的例子（即它可以让你多次输入＆＃39; C＆＃39;添加＆＃39;新宠物）

static String petname, petage;
    public static void main(String[] args) {
        initialText();
        String userinput = userInput();
        while (userinput.equals("A") || userinput.equals("B") || userinput.equals("C") || userinput.equals("D") || userinput.equals("E")) {
            if(userinput.equals("C")){
                System.out.println("Please type in a name");
                petname = userInput();
                System.out.println("Please type in an age");
                petage = userInput();
                Pet p = new Pet(petname, petage);
                System.out.println(petname + " has been added to the store.");
            }
            else{
                System.out.println("Option not configured yet");
                //TODO - the rest of the options
            }
            initialText();
            userinput = userInput();
        }
    }

    public static void initialText() {
        System.out.println("Welcome to the pet store.Type the letter to make your selection");
        System.out.println("A." + "  " + "List the pets in the store.");
        System.out.println("B." + "  " + "Age up the pets");
        System.out.println("C." + "  " + "Add a new pet");
        System.out.println("D." + "  " + "Adopt a pet");
        System.out.println("E." + "  " + "Quit");
    }

    public static String userInput(){
        Scanner s = new Scanner(System.in);
        return s.nextLine();
    }

它绝不是完美的，只是很快将它们拼凑在一起，让你有机会继续努力。

由于一些奇怪的原因被调用两次的方法？

1 个答案: