拜托,请帮帮我..
我正在开发一个项目,我从 JSON格式的Web服务获取数据.我试图解析它,但我无法做到.我有这个json数据 –
我正在开发一个项目,我从 JSON格式的Web服务获取数据.我试图解析它,但我无法做到.我有这个json数据 –
- {
- "response": {
- "status": {
- "code": "1","message": "sucess","user_id": "1"
- },"foods": [
- {
- "char": "A","content": [
- {
- "food_name": "add Malt"
- },{
- "food_name": "a la mode"
- },{
- "food_name": "Almonds"
- }
- ]
- },{
- "char": "Z","content": [
- {
- "food_name": "Zebra Cakes"
- },{
- "food_name": "Zucchini,Baby"
- },{
- "food_name": "zxc"
- }
- ]
- }
- ]
- }
- }
从这里我成功地获得了“食物”阵列,但当我试图获得“内容”数组和food_name数据时,我陷入困境.
我正在使用此代码,但我没有得到任何解决方案,请检查此剪辑代码.
- protected String doInBackground(String... args) {
- // Building Parameters
- List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
- nameValuePairs.add(new BasicNameValuePair("method","eat_tracking_details"));
- nameValuePairs.add(new BasicNameValuePair("uid",userid));
- // getting JSON string from URL
- JSONObject json = jsonParser.makeHttpRequest(JSONParser.urlname,"GET",nameValuePairs);
- //System.out.println("****json*"+json);
- if (json != null) {
- try {
- JSONObject response = json.getJSONObject("response");
- JSONObject status = response.getJSONObject("status");
- code = status.getString("code");
- JSONArray FoodArray = response.getJSONArray("foods");
- for (int i = 0; i < FoodArray.length(); i++) {
- String character = FoodArray.getJSONObject(i).getString("char");
- System.out.println("*****character****************"+character);
- JSONArray FoodNameArray = new JSONArray(FoodArray.getJSONObject(i).getString("content"));
- System.out.println("====================///////////"+FoodNameArray);
- for (int j = 0; j <FoodNameArray.length(); j++) {
- String Foodname = FoodArray.getJSONObject(j).getString("food_name");
- System.out.println("@@@@@@@@@@@@@"+Foodname);
- }
- }
- } catch (JSONException e) {
- // TODO: handle exception
- }
- }
检查此网址以获取网络服务响应 –
WEB-SERVICE URL