Question

我无法从HTTP请求中检索的JSON对象中检索所有键值。

这是从http请求返回的JSON对象：

 {"data":[{"movie_id":3,"movie_name":"Promithius","genre":"a dude","year":"2016","rating":45}]}

我的Android代码：

 try {
                //HttpJsonParser httpJsonParser = new HttpJsonParser();
                JSONObject JSonObj = (JSONObject)  new 
 JSONTokener(result).nextValue();
                String mymovie = JSonObj.getString("movie_name" );
                String movieGenre = JSonObj.getString("genre" );
                etResponse.setText("movie=" + mymovie + " genre=" + 
 movieGenre);

            } catch (JSONException e) {
                etResponse.setText("ERROR=" + e.getMessage());
                e.printStackTrace();
            }

当我仅在模拟器中运行它时：

 String mymovie = JSonObj.getString("movie_name" );
 etResponse.setText("movie=" + mymovie;

我没有错误地获得了电影名称。所以问题是我可以检索movie_name但没有流派。返回错误，提示“类型无价值” 预先感谢。

Answer 1

I think the data is JSONArray.

String result = "{\"data\":[{\"movie_id\":3,\"movie_name\":\"Promithius\",\"genre\":\"a dude\",\"year\":\"2016\",\"rating\":45}]}";

        try {
            JSONObject JSonObj = (JSONObject)  new JSONTokener(result).nextValue();
            JSONArray data = JSonObj.getJSONArray("data");

            for (int i=0; i<data.length(); i++){
                JSONObject jsonObject = data.getJSONObject(i);
                String mymovie = jsonObject.getString("movie_name" );
                String movieGenre = jsonObject.getString("genre" );
                String dest = "movie=" + mymovie + " genre=" + movieGenre;
                Log.d(TAG, dest);
            }

        } catch (JSONException e) {
            e.printStackTrace();
        }

Answer 2

下面是最终对我有用的代码。我希望它可以帮助其他人。

 try{
 JSONObject object = new JSONObject(result);
 JSONArray Jarray  = object.getJSONArray("data");
 object = Jarray.getJSONObject(0);
 String MovieName = object.getString("movie_name");
 String Genre = object.getString("genre");
 String Rating = object.getString("rating");
 String Year = object.getString("year");
 etResponse.setText("Movie Name = " + MovieName + " \n Genre = " + 
 Genre + "\n Rating = " + Rating + " \n Year = " + Year );
    } catch (JSONException e) {
    etResponse.setText("JSONException: " + e.getMessage());
    }

如何从JSON结果中检索流派

2 个答案: