把用戶輸入的某一點在數組索引

我試圖從HockeyPlayer構造函數獲取用戶輸入的HockeyPlayer類在數組「玩家」我HockeyPlayerMain類的陣列來存儲。每當他們按「A」它將調用構造函數，如果他們輸入數字＃6，我需要將對象/實例放入數組索引＃5，但我不知道如何做到這一點，非常初學數組。把用戶輸入的某一點在數組索引

對不起，如果這不明確，但我可以嘗試清除事情。

package hockeyplayer; 

import java.util.Scanner; 

public class HockeyPlayer { 

private String[] opponent = new String[10]; 
private int[] goalsScored = new int[10]; 
private int[] gameNumber = new int[10]; 
private String name; 
private int playerNumber; 
private int gameNum; 

public HockeyPlayer() { 

    Scanner input = new Scanner(System.in); 
    System.out.println("What is the name of the player?"); 
    name = input.nextLine(); 
    System.out.println("What is the player's number?"); 
    playerNumber = input.nextInt(); 
    input.nextLine(); 

} 


public String[] getOpponent() { 
    return opponent; 
} 

public void setOpponent(String[] opponent) { 
    this.opponent = opponent; 
} 

public int[] getGoalsScored() { 
    return goalsScored; 
} 

public void setGoalsScored(int[] goalsScored) { 
    this.goalsScored = goalsScored; 
} 

public String getName() { 
    return name; 
} 

public void setName(String name) { 
    this.name = name; 
} 

public int getPlayerNumber() { 
    return playerNumber; 
} 

public void setPlayerNumber(int playerNumber) { 
    this.playerNumber = playerNumber; 
} 

public void addGameDetails(){ 
    HockeyPlayer player1 = new HockeyPlayer(); 

    Scanner input = new Scanner(System.in); 
    System.out.println("What game number was it?"); 
    gameNumber[0] = input.nextInt(); 
    input.nextLine(); 
    System.out.println("Who were the opponents?"); 
    opponent[0] = input.nextLine(); 
    System.out.println("How many goals did the player score?"); 
    goalsScored[0] = input.nextInt(); 

} 

} 

package hockeyplayer; 
import java.util.Scanner; 

public class HockeyMain { 

private static String choice; 
private static HockeyPlayer[] players = new HockeyPlayer[12]; 
private static final String MENU = "Hockey Tracker\n"+ 
     "A-Add Player\n"+ 
     "G-Add game details\n"+ 
     "S-Show players\n"+ 
     "X-Quit\n"; 

public static void main(String[] args) { 

    Scanner input = new Scanner(System.in); 


do{  
    System.out.println(MENU); 
    choice = input.nextLine(); 

    switch(choice){ 
     case "A": 
      for(int i = 0; i < players.length; ++i){ 
      players[i] = new HockeyPlayer(); 
    } 
      break; 
     case "G": 

      break; 
     case "S": 
      break; 
     case "X":  
    } 

}while(!choice.equals("X")); 
} 



}

來源

2017-01-18 Shuckyducky

您的意思是'HockerPlayer'構造函數中給出的數字應該是'HockeyMain'類中數組播放器的索引嗎？ – nikowis

順便說一句，整個邏輯看起來很「奇怪」，但@Maciej Kowalski的回答是正確的。 – Vadim

在試試這個您的循環：

for(int i = 0; i < players.length; ++i){ 
     HockeyPlayer player = new HockeyPlayer();  
     players[player.getPlayerNmumber()] = player; 
    }

如果你想在「A」選項，按下一次進入只是一個球員，那麼你只需刪除for循環：

switch(choice){ 
    case "A": 
     HockeyPlayer p = new HockeyPlayer(); 
     players[p.getPlayerNumber()] = p; 
     break; 
    case "G": 

     break; 
    case "S": 
     break; 
    case "X":  
}

來源

2017-01-18 23:10:57

是的，這是我想要做的，謝謝。我早些時候嘗試過，但我可能會輸錯它。 – Shuckyducky

如果我不想一次完成12個，該怎麼辦？一次只有一個玩家，然後回到菜單？ – Shuckyducky

請看帖子。我已經加入該方案中 –

林不知道這是不是你的意思，但你可以在HockeyMain類使用getPlayerNumber()方法。如果用戶輸入不正確，這可能會導致數組覆蓋，這意味着數組將不會被完全填充。

do{  
System.out.println(MENU); 
choice = input.nextLine(); 

switch(choice){ 
    case "A": 
     for(int i = 0; i < players.length; ++i){ 
     HockeyPlayer p = new HockeyPlayer(); 
     players[p.getPlayerNumber()] = p; 
     } 
     break; 
    case "G": 

     break; 
    case "S": 
     break; 
    case "X":  
} 
}while(!choice.equals("X"));

來源

2017-01-18 23:15:25 nikowis

把用戶輸入的某一點在數組索引

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