JSON解析难题... 你碰到过没？？？

Question

从服务器上获取一个json的数据  请大家看看 应该怎么解析 PS：格式什么的都对，请不要纠结于括号上或者json的格式上！保证是对的！！！！ 听一个朋友说用foreach  但是不知道具体改怎么操作，有谁能指导下吗？？？？ "taxonomy":{"1":{"weight":"0","tid":"1","v_weight_unused":"0","vid":"1","description":"","name":"name1"}, 
"2":{"weight":"0","tid":"2","v_weight_unused":"0","vid":"1","description":"","name":"name2"}, 
"3":{"weight":"0","tid":"3","v_weight_unused":"0","vid":"1","description":"","name":"name3"}, 
"510":{"weight":"0","tid":"510","v_weight_unused":"0","vid":"1","description":"","name":"name4"}, 
"4":{"weight":"0","tid":"4","v_weight_unused":"0","vid":"5","description":"","name":"name5"}} 

Answer 1

把得到的字符串转换成JSONObject对象，再对该对象解析get("weight").toString()

Answer 2

测试通过,外面加了一层括号
StringBuilder jsonString = new StringBuilder();
            jsonString.append("{\"taxonomy\":{\"1\":{\"weight\":\"0\",\"tid\":\"1\",\"v_weight_unused\":\"0\",\"vid\":\"1\",\"description\":\"\",\"name\":\"name1\"},");
            jsonString.append("\"2\":{\"weight\":\"0\",\"tid\":\"2\",\"v_weight_unused\":\"0\",\"vid\":\"1\",\"description\":\"\",\"name\":\"name2\"},");
            jsonString.append("\"3\":{\"weight\":\"0\",\"tid\":\"3\",\"v_weight_unused\":\"0\",\"vid\":\"1\",\"description\":\"\",\"name\":\"name3\"},");
            jsonString.append("\"510\":{\"weight\":\"0\",\"tid\":\"510\",\"v_weight_unused\":\"0\",\"vid\":\"1\",\"description\":\"\",\"name\":\"name4\"},");
            jsonString.append("\"4\":{\"weight\":\"0\",\"tid\":\"4\",\"v_weight_unused\":\"0\",\"vid\":\"5\",\"description\":\"\",\"name\":\"name5\"} }}");            JSONObject taxonomy;
            taxonomy = new JSONObject(jsonString.toString());
            JSONArray names = taxonomy.names();
            String name;
            for (int y = 0; y < names.length(); y++) {
                name = (String) names.get(y);
                Log.e("taxonomy " + name, ""+taxonomy.getJSONObject(name));
            }

Answer 3

先格式化json,仔细看下其构成，如下：
根节点为taxonomy,包含5个子节点
每个子节点包含6个属性
"taxonomy":
{
"1":
{
"weight":"0",
"tid":"1",
"v_weight_unused":"0",
"vid":"1",
"description":"",
"name":"name1"
},  
"2":
{
"weight":"0",
"tid":"2",
"v_weight_unused":"0",
"vid":"1",
"description":"",
"name":"name2"
},  
"3":
{
"weight":"0",
"tid":"3",
"v_weight_unused":"0",
"vid":"1",
"description":"",
"name":"name3"
},  
"510":
{
"weight":"0",
"tid":"510",
"v_weight_unused":"0",
"vid":"1",
"description":"",
"name":"name4"
},  
"4":
{
"weight":"0",
"tid":"4",
"v_weight_unused":"0",
"vid":"5",
"description":"",
"name":"name5"
}
}

Answer 4

比如，要想得到  "tid":"510",这个值，代码如下：JSONObject json = new JSONObject(...);int tid510 = json.getJSONObject("taxonomy").getJSONObject("510").getInt("tid");

Answer 5

json可以得到子节点的名字列表的。

Answer 6

JsonArray + JsonObject.
不解释。

Answer 7

你是不是说的想使用递归方法自动循环遍历整个json数据？