如何使用fastJSON与JSON数组

本文关键字：JSON 数组 fastJSON 何使用 | 更新日期: 2023-09-27 18:13:20

我使用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;
    }
}

我不知道如何迭代数组数据与名称"敌人"

如何使用fastJSON与JSON数组

根据你目前编写的代码，你可以这样迭代敌人:

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

但是，我建议创建一些强类型的类来接收数据:

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似乎不能处理这个(我试过了，得到了一个异常)。但是，如果您切换到更健壮的库，如Json。Net，它工作起来没有问题:

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

这将允许您重写您的代码，以便您可以更轻松地处理您的数据:

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;
    }
}
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