當我已經聲明了變量時如何修復空錯誤

Question

我正在制作 PVP RPG 游戲，顯示框顯示為“null”，而不是我已經聲明的變量。

我已將該變量聲明為用戶的下一個輸入並將該信息存儲在該變量中。 然后當我嘗試顯示變量時，它只顯示“空”，

System.out.println("Welcome, Player One and Player Two!");
    delay(1500);
    System.out.println("What is your name, Player One?");
    playerOne.name = userInput.nextLine();

我已經將 playerOne 聲明為一個新角色（不同的類）

System.out.println("Your turn, " + playerOne.name+".");
    if (p1Swordgo == 1) {

這是我想出的問題。 它在同一個主方法中，變量在主方法中聲明，是的，我導入了掃描儀並聲明了變量 userInput

我希望它是用戶輸入的內容，但它提出了 null。 正如我之前所說的，它在同一個 main 方法中，應該不會出錯，但它出現了“null”

import java.util.Random;
import java.util.Scanner;


public class Arena {

Random generator = new Random();



public static void main(String[] args) {

    Character playerOne = new Character(10,10,0);
    Character playerTwo = new Character(10,10,0);

    boolean P1hasClass = false;
    boolean P2hasClass = false;
    int p1Swordgo = 0;
    int p2Alchgo = 0;
    int p2Archgo = 0;


    Scanner userInput = new Scanner(System.in);

System.out.println("Welcome, Player One and Player Two!");
    delay(1500);
    System.out.println("What is your name, Player One?");
    playerOne.name = userInput.nextLine();
    delay(1000);
    System.out.println("Hello, " +playerOne.name +".");
    delay(1000);
    System.out.println("What is your name, Player Two?");
    playerTwo.name = userInput.nextLine();
    delay(1000);
    System.out.println("Hello, " +playerTwo.name +".");
    delay(1500);

countdown();

        System.out.println("Your turn, " + playerOne.name+".");
    if (p1Swordgo == 1) {
        if (p2Archgo == 1 || p2Alchgo == 1) {
            if (playerOne.move == 1){

                System.out.println("What do you want to do?" +'\n' +"1 = Move into range of " +playerTwo.name +'\n' +"2 = Heal" +'\n' +"3 = Forfeit");
                int P1Choice = userInput.nextInt();
                if (P1Choice == 1) {
                    playerOne.move --;
                    System.out.println(playerOne.move);
                }

            }



        }

    }





}
















public static void delay ( int time){
    try {
        Thread.sleep(time);
    } catch (InterruptedException ex) {
        Thread.currentThread().interrupt();
    }
}

public static void countdown() {
    delay(500);
    System.out.println("Get ready to fight in 5,");
    delay(1000);
    System.out.println("4");
    delay(1000);
    System.out.println("3");
    delay(1000);
    System.out.println("2");
    delay(1000);
    System.out.println("1");
    delay(1000);
    System.out.println("Fight!");
    delay(750);
}

}

然后在一個叫做 Character 的類中

public class Character {


public int strength;
public double health;
public int move;
public String name;


public Character(double health, int strength, int move) {

    this.health = health;
    this.strength = strength;
    this.name = name;
    this.move = move;

}

}

在一個叫做SwordFighter的課程中

public class SwordFighter extends Character {

public SwordFighter() {
    super(60,15, 1);
}

}

在一個叫做 Archer 的班級里

public class Archer extends Character{

public Archer() {
    super(45,20, 0);
}

}

最后，在一個名為 Alchemist public class Alchemist extends Character {

public Alchemist() {
    super(50,15, 0);
}

}

順便說一下，感謝您的耐心等待

Answer 1

一旦兩個玩家選擇了他們的名字並且你已經使用playerOne.name = userInput.nextLine();設置了它playerOne.name = userInput.nextLine(); ，您分配一個不同的對象，名稱為空，給playerOne ：

 playerOne = new SwordFighter();

因此，在執行此行后， playerOne.name為空。