记忆游戏 - 如何匹配不同的图像？

Question

我知道这听起来可能很奇怪，但我需要制作一个记忆游戏来匹配不同的图像而不是两个相同的图像，例如，第一和伸出食指的手

这是我匹配两个相同图像的实际代码，它克隆了使相同图像出现两次的卡片

下面的场景控制器

using System.Collections;
using System.Collections.Generic;
using UnityEngine;
using UnityEngine.SceneManagement;

public class SceneController : MonoBehaviour {

public const int gridRows = 2;
public const int gridCols = 5;
public const float offsetX = 4f;
public const float offsetY = 5f;

[SerializeField] private MainCard originalCard;
[SerializeField] private Sprite[] images;

private void Start()
{
    Vector3 startPos = originalCard.transform.position; //The first position of the cards, all other cards are based on this one

    int[] numbers = { 0, 0, 1, 1, 2, 2, 3, 3, 4, 4};
    numbers = ShuffleArray(numbers); 

    for(int i = 0; i < gridCols; i++)
    {
        for(int j = 0; j < gridRows; j++)
        {
            MainCard card;
            if(i == 0 && j == 0)
            {
                card = originalCard;
            }
            else
            {
                card = Instantiate(originalCard) as MainCard;
            }

            int index = j * gridCols + i;
            int id = numbers[index];
            card.ChangeSprite(id, images[id]);

            float posX = (offsetX * i) + startPos.x;
            float posY = (offsetY * j) + startPos.y;
            card.transform.position = new Vector3(posX, posY, startPos.z);
        }
    }
}

private int[] ShuffleArray(int[] numbers)
{
    int[] newArray = numbers.Clone() as int[];
    for(int i = 0; i < newArray.Length; i++)
    {
        int tmp = newArray[i];
        int r = Random.Range(i, newArray.Length);
        newArray[i] = newArray[r];
        newArray[r] = tmp;
    }
    return newArray;
}

//-------------------------------------------------------------------------------------------------------------------------------------------

private MainCard _firstRevealed;
private MainCard _secondRevealed;

private int _match = 5;
[SerializeField] private TextMesh matchLabel;

public bool canReveal
{
    get { return _secondRevealed == null; }
}

public void CardRevealed(MainCard card)
{
    if(_firstRevealed == null)
    {
        _firstRevealed = card;
    }
    else
    {
        _secondRevealed = card;
        StartCoroutine(CheckMatch());
    }
}

private IEnumerator CheckMatch()
{
    if(_firstRevealed.id == _secondRevealed.id)
    {
        _match--;
        matchLabel.text = "Pares Restantes: " + _match;
        if(_match == 1){
            matchLabel.text = "Somente " + _match + " Par Restante";
        }
        if(_match == 0){
            SceneManager.LoadScene("Sucesso");
        }
    }
    else
    {
        yield return new WaitForSeconds(0.5f);

        _firstRevealed.Unreveal();
        _secondRevealed.Unreveal();
    }

    _firstRevealed = null;
    _secondRevealed = null;

}

public void Menu()
{
    SceneManager.LoadScene("Menu");
}

public void Fases()
{
    SceneManager.LoadScene("Fases");
}

public void Sair()
{
    Application.Quit ();
}

public void Level1()
{
    SceneManager.LoadScene("Level 1");
}

public void Sucesso()
{
    SceneManager.LoadScene("Sucesso");
}

}

下面的主卡类

using System.Collections;
using System.Collections.Generic;
using UnityEngine;

public class MainCard : MonoBehaviour {

[SerializeField] private SceneController controller;
[SerializeField] private GameObject Card_Back;


public void OnMouseDown()
{
    if(Card_Back.activeSelf && controller.canReveal)
    {
        Card_Back.SetActive(false);
        controller.CardRevealed(this);
    }
}


private int _id;
public int id
{
    get { return _id; }
}

public void ChangeSprite(int id, Sprite image)
{
    _id = id;
    GetComponent<SpriteRenderer>().sprite = image; //Isso pega o Sprite Renderer e muda a propriedade do Sprite
}

public void Unreveal()
{
    Card_Back.SetActive(true);
}

}

PS：对不起我的英语

Answer 1

您可以像这个简单的例子一样将每个图像封装在一个类的对象中：

class CardImage : MonoBehaviour
{
    public Sprite cardSprite;
    public int cardNumber;
}

然后在你的游戏逻辑中，你可以像这样检查 cardNumber 字段来比较两张牌：

if (card1.cardNumber == card2.cardNumber)
{
// Do what you have to do when two cards are equal
}

有关 OOP 和封装的更多信息，请参阅 this