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我正在使用fastJSON从我制作的 JSON 文件中读取数据（JSON 文件包含 Unity 项目的游戏关卡数据，但这并不重要）。这是 JSON 内容：

{"1": {
    "background": "background1.png",
    "description": "A description of this level",
    "enemies": 
              [{"name": "enemy1", "number": "5"},
               {"name": "enemy2", "number": "2"}]},
 "2": {
    "background": "background1.png",
    "description": "A description of this level",
    "enemies": 
              [{"name": "enemy1", "number": "8"},
               {"name": "enemy2", "number": "3"}]},
"3": {
    "background": "background2.png",
    "description": "A description of this level",
    "enemies": 
              [{"name": "enemy2", "number": "5"},
               {"name": "enemy3", "number": "3"},
               {"name": "enemy4", "number": "1"}]}
}

这是我的代码：

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

public class LevelManager : MonoBehaviour {

    private Dictionary<string, object> currentLevelData;

    public TextAsset levelJSON;
    public int currentLevel;

    // Use this for initialization
    void Start () {

        currentLevelData = LevelElements (currentLevel);
        if (currentLevelData != null) {
            Debug.Log (currentLevelData["background"]);
            Debug.Log (currentLevelData["description"]);
            /* How iterate the "enemies" array */
        } else {
            Debug.Log ("Could not find level '" + currentLevel + "' data");
        }
    }

    Dictionary<string, object> LevelElements (int level) {
        string jsonText = levelJSON.ToString();
        Dictionary<string, object> dictionary = fastJSON.JSON.Parse(jsonText) as Dictionary<string, object>;

        Dictionary<string, object> levelData = null;
        if (dictionary.ContainsKey (level.ToString ())) {
            levelData = dictionary [level.ToString ()] as Dictionary<string, object>;
        }

        return levelData;
    }
}

我不知道如何用名称“敌人”迭代数组数据。

4

1 回答 1

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使用您当前编写代码的方式，您将像这样遍历敌人：

foreach (Dictionary<string, object> enemy in (List<object>)currentLevelData["enemies"])
{
    Debug.Log(enemy["name"]);
    Debug.Log(enemy["number"]);
}

但是，我建议制作一些强类型类来接收您的数据：

public class Level
{
    public string background { get; set; }
    public string description { get; set; }
    public List<Enemy> enemies { get; set; }
}

public class Enemy
{
    public string name { get; set; }
    public string number { get; set; }
}

理想情况下，这将允许您像这样反序列化：

Dictionary<string, Level> dictionary = 
                          fastJSON.JSON.ToObject<Dictionary<string, Level>>(jsonText);

不幸的是，fastJSON 似乎无法处理这个问题（我试过了，但遇到了异常）。但是，如果您切换到更健壮的库，例如Json.Net，它可以正常工作：

Dictionary<string, Level> dictionary = 
                        JsonConvert.DeserializeObject<Dictionary<string, Level>>(jsonText);

这将允许您重写代码，以便更轻松地处理数据：

public class LevelManager : MonoBehaviour
{
    private Level currentLevelData;

    public string levelJSON;
    public int currentLevel;

    // Use this for initialization
    void Start()
    {
        currentLevelData = LevelElements(currentLevel);
        if (currentLevelData != null)
        {
            Debug.Log(currentLevelData.background);
            Debug.Log(currentLevelData.description);

            foreach (Enemy enemy in currentLevelData.enemies)
            {
                Debug.Log(enemy.name);
                Debug.Log(enemy.number);
            }
        }
        else
        {
            Debug.Log("Could not find level '" + currentLevel + "' data");
        }
    }

    Level LevelElements(int level)
    {
        string jsonText = levelJSON.ToString();
        Dictionary<string, Level> dictionary = 
                JsonConvert.DeserializeObject<Dictionary<string, Level>>(jsonText);

        Level levelData = null;
        if (dictionary.ContainsKey(level.ToString()))
        {
            levelData = dictionary[level.ToString()];
        }

        return levelData;
    }
}
于 2016-10-16T19:26:04.050 回答