如何使用fastJSON与JSON数组
如何使用fastJSON与JSON数组
提问于 2016-10-16 16:02:41
我正在使用fastJSON从我创建的JSON文件中读取数据( JSON文件中包含一个游戏级别的数据，用于Unity的项目，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"}]}
}这是我的密码：
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“敌人”的名字来迭代数组数据。
回答 1
Stack Overflow用户
回答已采纳
发布于 2016-10-16 19:26:04
按照当前的代码编写方式，您可以像这样迭代敌人：
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似乎无法处理这个问题(我尝试过了，得到了一个异常)。但是，如果切换到一个更健壮的库(如Json.Net )，它可以正常工作：
Dictionary<string, Level> dictionary =
JsonConvert.DeserializeObject<Dictionary<string, Level>>(jsonText);这将使您可以重写代码，以便更容易地处理数据：
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;
}
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/40072313
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