使用對象沒有構造函數參數的JAVA

Question

所以基本上我創建一個投注計劃，使PLAYER1，Player2，Player3等都是對象，並各自具有的-1 5個初始化場，這意味着沒有賭，但。

我要求用戶輸入他們的下注數，（這些數字數組內的整數），他們可以有最多的五注，解釋-1五個初始化領域。

這裏的問題是我似乎不能找到一種方法，以具有一定的玩家只輸入1倍或2的投注，而另一個輸入端4和5的另一例如。我的程序強制每個用戶輸入5個整數，即使他們拒絕下注5次。 任何方式，我可以使這項工作？

因此，這裏是我的驅動程序類：（它的另一部分是無關緊要的）

System.out.println("Wheel : 0-32-15-19-4-21-2-25-17-34-6-27-13-36-11-30-8-23-10-5-24-16-33-1-20-14-31-9-22-18-29-7-28-12-35-3-26"); 
     System.out.println("Dealer…Place your bets"); 

     // initializing the six players of the game 

     Player player1 = new Player(-1,-1,-1,-1,-1); 
     System.out.println("Player 1: "); 
      int number1 = keyin.nextInt(); 
      int number2 = keyin.nextInt(); 
      int number3 = keyin.nextInt(); 
      int number4 = keyin.nextInt(); 
      int number5 = keyin.nextInt(); 
       System.out.println(""); 

     Player player2 = new Player(-1, -1, -1, -1, -1); 
     System.out.println("Player 2: "); 
      number1 = keyin.nextInt(); 
      number2 = keyin.nextInt(); 
      number3 = keyin.nextInt(); 
      number4 = keyin.nextInt(); 
      number5 = keyin.nextInt(); 
       System.out.println(""); 

     Player player3 = new Player(-1, -1, -1, -1, -1); 
     System.out.println("Player 3: "); 
      number1 = keyin.nextInt(); 
      number2 = keyin.nextInt(); 
      number3 = keyin.nextInt(); 
      number4 = keyin.nextInt(); 
      number5 = keyin.nextInt(); 
       System.out.println(""); 

     Player player4 = new Player(-1, -1, -1, -1, -1); 
      System.out.println("Player 4: "); 
      number1 = keyin.nextInt(); 
      number2 = keyin.nextInt(); 
      number3 = keyin.nextInt(); 
      number4 = keyin.nextInt(); 
      number5 = keyin.nextInt(); 
       System.out.println(""); 

     Player player5 = new Player(-1, -1, -1, -1, -1); 
      System.out.println("Player 5: "); 
      number1 = keyin.nextInt(); 
      number2 = keyin.nextInt(); 
      number3 = keyin.nextInt(); 
      number4 = keyin.nextInt(); 
      number5 = keyin.nextInt(); 
       System.out.println(""); 

     Player player6 = new Player(-1, -1, -1, -1, -1); 
      System.out.println("Player 6: "); 
      number1 = keyin.nextInt(); 
      number2 = keyin.nextInt(); 
      number3 = keyin.nextInt(); 
      number4 = keyin.nextInt(); 
      number5 = keyin.nextInt(); 
       System.out.println(""); 

這裏是我的球員其他類：

public class Player { 

int number1; 
int number2; 
int number3; 
int number4; 
int number5; 
// constructor for each player at the table 
public Player(int number1, int number2, int number3, int number4, int number5) { 

    this.number1 = number1; 
    this.number2 = number2; 
    this.number3 = number3; 
    this.number4 = number4; 
    this.number5 = number5; 

的數字1，數字代表...五個賭注。

而且澄清，用戶可以不被要求輸入某些字符串，整型，任何字符，讓程序知道他/她做輸入他們的賭注。它看起來像這樣

Player 1: 12 13 15 
Player 2: 0 5 4 

Answer 1

跟進你的評論Patrick。 這是你如何使用Integer.parseInt

...  
Player player2 = new Player(-1, -1, -1, -1, -1); 
System.out.println("Player 2: "); 
try { 
       number1 = Integer.parseInt(keyin.nextLine()); 
       number2 = Integer.parseInt(keyin.nextLine()); 
       number3 = Integer.parseInt(keyin.nextLine()); 
       number4 = Integer.parseInt(keyin.nextLine()); 
       number5 = Integer.parseInt(keyin.nextLine()); 
       System.out.println(""); 
    } 
    //Code will go to this block when user entered something other than integers 
    catch(NumberFormatException e) { 
     System.out.println("Done accepting bets from player2"); 
    } 
    ... 

另外，可能的keyin.nextLine()輸出存入一個字符串，並應用parseInt函數或評估它的一些「關鍵詞」你想，像這樣。

String keyedinString = keyin.nextLine(); 
if(keyinString.equalsIgnoreCase("done betting") { 
    throw new Exception(); 
} 

你很快就會意識到這個過程是多麼的重複。因此，正如其他人所說的，建議考慮使用數組和循環來完成任務。像這樣。

int[] player1Bets = new int[5]; 
for(int i=0;i<5;i++) { 
    try { 
     player1Bets[i] = Integer.parseInt(keyin.nextLine()); 
    } 
    catch(NumberFormatException e) { 
     System.out.println("Player1 took" + i + "bets. he is done betting"); 
    } 
} 
... 

這樣，而不是number1, number2, number3, number4 ...等你得的第一注中player1Bets[0], second bet in player1Bets[1] ......更容易打出來，和代碼。

希望事情很清楚。

Answer 2

有很多方法，你可以這樣做：

• 使用1,2,3,4,5參數的構造函數，而不是隻要有一個5參數。
• 更好的方法是使用BUilder模式，因爲稍後您可能需要放置超過5次的投注，所以不要繼續添加構造函數，這是更好的選擇。
• 您也可以使用可變參數。見When do you use varargs in Java?
• 就像@immibs說，你也可以有像一個成員變量列表，然後您可以添加到你想要的

Answer 3

您可以隨時循環輸入，並指定一定數量的循環，以儘可能多終止。例如;投注爲1,2和3，並且從那裏停止輸入，用戶可以輸入0然後退出循環，或者程序可以每次詢問用戶他們是否想要再次投注。例如，

下注：1

是否要再次下注？按1，不是和0爲不

如果這個問題更多的是對邏輯的一部分，那麼這可能會幫助你。我希望我有所幫助，我是新來者，我試着回答問題以提高我的技能和經驗。 :)

Answer 4

如果您想傳遞多個參數並且參數計數不固定，則您可能需要「構建模式」。請參閱https://jlordiales.wordpress.com/2012/12/13/the-builder-pattern-in-practice/瞭解更多關於「Builder Pattern」的信息。

下面是一個例子：

如果你有一個User類有多個參數，

public class User { 
    private final String firstName; // required 
    private final String lastName; // required 
    private final int age; // optional 
    private final String phone; // optional 
    private final String address; // optional 

    public static class UserBuilder { 
    private final String firstName; 
    private final String lastName; 
    private int age; 
    private String phone; 
    private String address; 
    public UserBuilder(String firstName, String lastName) { 
     this.firstName = firstName; 
     this.lastName = lastName; 
    } 
    public UserBuilder withAge(int age) { 
     this.age = age; 
     return this; 
    } 
    public UserBuilder withPhone(String phone) { 
     this.phone = phone; 
     return this; 
    } 
    public UserBuilder withAddress(String address) { 
     this.address = address; 
     return this; 
    } 
    public User build() { 
     // if(age < 0) { ... } return new User(this); 
     // not thread-safe, a second thread may modify the value of age 
     User user = new User(this); 
     if (user.getAge() < 0) { 
     throw new IllegalStateException("Age out of range!"); // thread-safe 
     } 
     return user; 
    } 
    } 

    private User(UserBuilder builder) { 
    this.firstName = builder.firstName; 
    this.lastName = builder.lastName; 
    this.age = builder.age; 
    this.phone = builder.phone; 
    this.address = builder.address; 
    } 

    public String getFirstName() { 
    return firstName; 
    } 
    public String getLastName() { 
    return lastName; 
    } 
    public int getAge() { 
    return age; 
    } 
    public String getPhone() { 
    return phone; 
    } 
    public String getAddress() { 
    return address; 
    } 
} 

使用下面的方法來建立一個新的User實例：

User user1 = new User.UserBuilder("Jhon", "Doe") 
        .withAge(30) 
        .withPhone("1234567") 
        .withAddress("Fake address 1234") 
        // maybe more parameters 
        .build(); 
User user2 = new User.UserBuilder("Tom", "Jerry") 
        .withAge(28) 
        // maybe more parameters 
        .build();