Libgdx - 如何从数组中获取特定项？

I have a Player class which has

我有一个Player类

int number;

In main, I store them in Array.

在main中,我将它们存储在Array中。

Array<Player> players;

How can I get a player which has for example number=2?

如何获得一个数字= 2的玩家?

3 个解决方案

#1

This question is programming in general, and has several ways to do what you say, I recommend you look for OOP.

这个问题一般是编程,有几种方法可以做你说的,我建议你寻找OOP。

a solution, general, would encapsulate the variable, "creating getter and setter, you need, eg:

一般来说,解决方案将封装变量,“创建getter和setter,你需要,例如:

this a simple class;

这个简单的课;

private int number;

public Player(int num){
   this.number = num;
}

public int getNumber (){
   return number;
}

public void setNumber (int n){
   this.number = n;
}

. In your code for search you can use the solution, using a for, or an iterator

。在您的搜索代码中,您可以使用解决方案,使用for或迭代器

for (int a = 0; a < players.size(); a++){

    int tmpNumber = players.get(a).getNumber();

    if(tmpNumber == 2){
      players.get(a); //the object array with index equal to 
                      //the value of 'a', have, number 2 stored 
                      //in the variable 'number'
    }
}

but this question I think is a little matter of taste or needs you have

但我认为这个问题只是一个品味或需求的问题

#2

I'm going to make some assumptions to demonstrate it:

我将做一些假设来证明它:

Let's assume you have a constructor in your Player class:

假设你的Player类中有一个构造函数:

public Player(int num){
   this.number = num;
}

And you have added some players in your arraylist:

你已经在你的arraylist中添加了一些玩家:

players.add(new Player(1));
players.add(new Player(2));
players.add(new Player(3));

Now loop through your array and find the one where number == 2

现在循环遍历您的数组并找到number == 2的数组

for (int i = 0; i < players.size(); i++){
    Player tmp = players.get(i);
    if(tmp.number == 2){
        System.out.println("Index of player number 2: " + players.indexOf(tmp));
    }
}

#3

I realised that I'd misunderstood the question with my previous answer. Here's a better one...

我意识到我用我以前的答案误解了这个问题。这是一个更好的...

Those recommending you iterate through the Array are correct, but if you have a lot of Players this will be ineffeicient. In that case, it would be better to use a Map where the key was the player number, and the value is the player itself.

那些推荐你迭代数组的人是正确的,但如果你有很多玩家,这将是无效的。在这种情况下,最好使用一个Map,其中键是玩家编号,值是玩家本身。

Here's some sample code showing the approach (most of it is just setting up some test data)...

这里有一些示例代码显示了该方法(大多数只是设置一些测试数据)......

import java.util.HashMap;
import java.util.Map;

public class Temp {

    public static void main(String[] args) {

        // Create a map instead of an array
        Map<Integer, Player> players = new HashMap<>();

        // Quick hack just to put some data in it.
        for(int i = 0; i < 1000; i++) {
            players.put(i, new Player(i));
        }

        // Once they're in a map, retrieving them is simple...
        Player playerFromMap = players.get(537);

        // Check we got it right
        System.out.println(playerFromMap.getNumber()); // 537
    }

    private static class Player {

        private int number;

        private Player(int number) {
            this.number = number;
        }

        public int getNumber() {
            return number;
        }
    }
}

#1