Json解析，新手求帮助

[small-dream]

服务端返回的Json字符串如下

我应该怎么才能把它去取出来，里面有嵌套，怎么办？

[编程点滴]

Json.getJson()就可以获取到一个json对象啊，这样就可以处理嵌套了

[teemai]

用JSONObject解析好了

[teemai]

参考

JSONArray jsonArrays = new JSONArray(buf);
			for (int i = 0; i < jsonArrays.length(); i++) {
				JSONObject obj = jsonArrays.getJSONObject(i);
				long count = 0L;
				String categoryType = "";
				String catTypeToString = "";
				String url = "";
				if (obj.has("count")) {
					count = Long.valueOf(obj.optString("count"));
				}
				
				if (obj.has("cateType")) {
					categoryType = obj.optString("cateType");
				}
				
				String cateName = null;
				if (obj.has("cateName")) {
					cateName = obj.optString("cateName");
				}
				url = obj.optString("url");
				if (QuhaoConstant.test) {
					if(null != url &&!"".equals(url))
					{
						url = obj.getString("url").replace("http://localhost:9081/", QuhaoConstant.HTTP_URL);
					}
					
//					imgUrl = obj.getString("merchantImage").replace("http://localhost:9081/", "http://192.168.2.100:9081/");
					System.out.println(url);
				} else {
					url = obj.getString("merchantImage");
				}
				Category category = new Category(count, categoryType, cateName, url);
				categroys.add(category);
			}

[代码间的舞者]

好像用到eval eval(“【返回的对象】”)

[showjim]

试试fastCSharp

        class result
        {
            public bool ValidResult;
            public userInfo ResultUserLoginInfo;
            //...
        }
        class userInfo
        {
            public int UserId;
            //...
        }
            result result = fastCSharp.setup.cSharp.json.parse.Get<result>(json);

[tony4geek]

循环解析 json

[leokelly001]

JSONObject解析

[代码间的舞者]

引用 3 楼 huxiweng 的回复:

参考

JSONArray jsonArrays = new JSONArray(buf);
			for (int i = 0; i < jsonArrays.length(); i++) {
				JSONObject obj = jsonArrays.getJSONObject(i);
				long count = 0L;
				String categoryType = "";
				String catTypeToString = "";
				String url = "";
				if (obj.has("count")) {
					count = Long.valueOf(obj.optString("count"));
				}
				
				if (obj.has("cateType")) {
					categoryType = obj.optString("cateType");
				}
				
				String cateName = null;
				if (obj.has("cateName")) {
					cateName = obj.optString("cateName");
				}
				url = obj.optString("url");
				if (QuhaoConstant.test) {
					if(null != url &&!"".equals(url))
					{
						url = obj.getString("url").replace("http://localhost:9081/", QuhaoConstant.HTTP_URL);
					}
					
//					imgUrl = obj.getString("merchantImage").replace("http://localhost:9081/", "http://192.168.2.100:9081/");
					System.out.println(url);
				} else {
					url = obj.getString("merchantImage");
				}
				Category category = new Category(count, categoryType, cateName, url);
				categroys.add(category);
			}

版主，楼主是想要客户端，即js解析JSon的代码吧

[huntor]

可以使用 json-path ,类似xpath的方式 http://code.google.com/p/json-path/

[悲鸣秋]

我一般用$.getJSON("调用方法",map,function(data){ //然后 你想调用 ResultUserInfo var resultUserInfo = data.ResultUserInfo; //接着要调用里面的 如UserName var userName = resultUserInfo.UserName; })