如何在 Unity 游戏中将游戏设置保存到文件中而不返回序列化错误？

Question

我正在构建一个 Android 游戏，“选项”菜单提供 3 种不同的设置：背景音乐音量、音效音量和射击按钮的屏幕侧位置。还有应该保存的高分。

我尝试根据某人提供的答案创建脚本，但是当我使用音乐音量滑块对其进行测试时，它不断返回有关无法序列化的内容的错误。起初它说“SerializationException: Type 'GameData' in Assembly 'Assembly-CSharp, Version=0.0.0.0, Culture=neutral, PublicKeyToken=null' 未标记为可序列化”，所以我尝试添加“[System.Serializable]”在某处建议的类代码上方，但现在它返回错误“SerializationException: Type 'UnityEngine.MonoBehaviour' in Assembly 'UnityEngine.CoreModule, Version=0.0.0.0, Culture=neutral, PublicKeyToken=null' is notmarked as serializable ”。

此外，在序列化错误之后，它还会写入错误“IOException：在路径 C:\Users\UserName\AppData\LocalLow\CompanyName\AppName\save.dat 上共享冲突”。

代码如下：

using System.Collections;
using System.Collections.Generic;
using UnityEngine;
using System.IO;
using System.Runtime.Serialization.Formatters.Binary;
using System.Security.Cryptography;

[System.Serializable]
public class GameData : MonoBehaviour
{
    public int highScore = 0;
    public float musicVol = 1;
    public float sfxVol = 1;
    public int shootSide = 1;
    // Start is called before the first frame update
    private void Start()
    {
        LoadFile();
    }
    public void HighScoreValue(int highS)
    {
        highScore = highS;
        SaveData();
    }
    public void MusicValue(float musicV)
    {
        musicVol = musicV;
        SaveData();
    }
    public void SfxValue(float sfxV)
    {
        sfxVol = sfxV;
        SaveData();
    }
    public void ShootSideValue(int shootS)
    {
        shootSide = shootS;
        SaveData();
    }
    private void SaveData()
    {
        string destination = Application.persistentDataPath + "/save.dat";
        FileStream file;

        if (File.Exists(destination))
        {
            file = File.OpenWrite(destination);
        }
        else
        {
            file = File.Create(destination);
        }
        GameData data = gameObject.AddComponent<GameData>();
        BinaryFormatter formatter = new BinaryFormatter();
        formatter.Serialize(file, data);
        file.Close();
    }
    private void LoadFile()
    {
        string destination = Application.persistentDataPath + "/save.dat";
        FileStream file;

        if (File.Exists(destination))
        {
            file = File.OpenRead(destination);
        }
        else
        {
            Debug.Log("File not found");
            return;
        }
        BinaryFormatter formatter = new BinaryFormatter();
        GameData data = (GameData)formatter.Deserialize(file);
        file.Close();
        highScore = data.highScore;
        musicVol = data.musicVol;
        sfxVol = data.sfxVol;
        shootSide = data.shootSide;
    }
}

代码有什么问题？提前致谢。

Answer 1

MonoBehaviour 可能无法使用反序列化程序所需的 new 关键字来实例化。不需要额外的 AddComponent 和 GetComponent，因为此脚本已经是您要在 GameObject 上使用的组件。

---

您应该（至少我会）将可序列化类型分离为纯数据类/结构，并在实现逻辑的 MonoBehaviour 中使用它的实例。

可能看起来像例如

[Serializable]
public class GameDataContainer
{
    public int highScore = 0;
    public float musicVol = 1;
    public float sfxVol = 1;
    public int shootSide = 1;
}

public class GameData : MonoBehaviour
{
    // since this is a serialized, 
    // it will be initialized with a valid instance by default
    public GameDataContainer data;

    // NEVER use +"/" for system paths!
    // initilaize this only once
    private string destination => Path.Combine(Application.persistentDataPath, "save.dat");

    // Start is called before the first frame update
    private void Start()
    {
        LoadFile();
    }

    public void HighScoreValue(int highS)
    {
        data.highScore = highS;
        SaveData();
    }

    public void MusicValue(float musicV)
    {
        data.musicVol = musicV;
        SaveData();
    }

    public void SfxValue(float sfxV)
    {
        data.sfxVol = sfxV;
        SaveData();
    }

    public void ShootSideValue(int shootS)
    {
        data.shootSide = shootS;
        SaveData();
    }

    private void SaveData()
    {
        using (var file = File.Open(destination, FileMode.Create, FileAccess.Write, FileShare.Write))
        {
            var formatter = new BinaryFormatter();
            formatter.Serialize(file, data);
        }
    }

    private void LoadFile()
    {
        if (!File.Exists(destination))
        {
            Debug.Log("File not found");
            return;
        }

        using(var file = File.Open(destination, FileMode.Open, FileAccess.Read, FileShare.Read))
        {
            var formatter = new BinaryFormatter();
            data = (GameDataContainer)formatter.Deserialize(file);
        }
    }
}

Answer 2

由于您的 GameData 类继承自 Unity 的 MonoBehaviour，因此序列化需要 MonoBehaviour 也可序列化；它似乎不是。

也许这会有所帮助：Serialize & Deserialize Unity3D MonoBehaviour script