Question

我正在尝试将列表保存到XML中，稍后将 XML重新加载到该列表中。除了实际的Save方法之外，其中大多数似乎都能正常工作。

这是我的代码。问题是没有创建 SaveInformation.xml 。

[XmlRoot("SaveInformation")]
public class SaveInformation {

    [XmlArray("stat")]
    [XmlArrayItem("PlayerData")]
    public PlayerData[] stat;

    public void Save(string path){
        var serializer = new XmlSerializer(typeof(SaveInformation));
        using(var stream = new FileStream(path,FileMode.Create)){
            serializer.Serialize (stream, this);
            vstream.Close();
        }
    }

    public static SaveInformation Load(string path){
        var serializer = new XmlSerializer(typeof(SaveInformation));
        using(var stream = new FileStream(path,FileMode.Open)){
            return serializer.Deserialize (stream) as SaveInformation;
            stream.Close();
        }
    }
}

以下是Load和Save方法：

using UnityEngine;
using System.Collections;
using System.IO;

public class StoreData : MonoBehaviour {

    public static void Load(){
        var saveInformation = SaveInformation.Load(
            Path.Combine(Application.dataPath, "SaveInformation.xml"));

        Debug.Log("Loaded");
    }

    public static void Save(){
        SaveInformation obj = new SaveInformation();

        obj.Save(Path.Combine(Application.persistentDataPath,
            "SaveInformation.xml"));

        Debug.Log("Saved");
    }
}

以下是应该传递XML的类 stat 和 PlayerData

using UnityEngine;
using System.Collections;
using System.Collections.Generic;
using System.Xml;
using System.Xml.Serialization;

[XmlRoot("PlayerDatabaseCollection")]
public class PlayerDatabase : MonoBehaviour {
    [XmlArray("stat")]
    [XmlArrayItem("PlayerData")]
    public List<PlayerData> stat = new List<PlayerData>();
}

这是数据本身

using UnityEngine;
using System.Collections;
using System.Xml;
using System.Xml.Serialization;

[System.Serializable]
public class PlayerData {
    [XmlAttribute("name")]
    public string Name;
    [XmlAttribute("current")]
    public int Current;
    [XmlAttribute("max")]
    public int Max;
}

Answer 1

如果您调用save，则不应创建类的新实例。确保您只有一个获得加载和保存的实例...

public class StoreData : MonoBehaviour {

    // this will hold your data (but in the snippets you provided I 
    // can't see how you are accessing this, maybe this should be public)
    private static SaveInformation saveInformation = null; // keep one instance

    private static string savepath = Application.dataPath; // or Application.persistentDataPath ?
    private static string savePathFile = Path.Combine(savepath, "SaveInformation.xml");

    public SaveInformation Instance {
       get {
           return (saveInformation ?? new SaveInformation());
       }
    }

    public static void Load(){
        saveInformation = SaveInformation.Load(savePathFile);
        Debug.Log("Loaded");
    }

    public static void Save(){
        if (saveInformation != null) {
           saveInformation.Save(savePathFile);
           Debug.Log("Saved");
        } else {
            Debug.Log("No saveinformation yet");
        }
    }
}

XML文件无法使用Unity和C＃编写

1 个答案: