Question

我正在尝试开发一个快速的应用程序，它将从这个网站返回即将上映的电影及其精简演员：http://developer.rottentomatoes.com/docs/read/json/v10/Upcoming_Movies。但是，我不确定如何访问最新的电影以及删节演员。如果有人可以帮助我，我会非常感激，因为JSON对我来说有点困惑。

到目前为止，我已经使用How to request movie data from Rotten Tomatoes using their JSON API?来帮助我，但我希望能够以“The Hangover：Bradley Cooper等”的格式列出电影和演员阵容。到目前为止是这个

if (response != null)
{
    try
    {
        // convert the String response to a JSON object,
        // because JSON is the response format Rotten Tomatoes uses
        JSONObject jsonResponse = new JSONObject(response);

        // fetch the array of movies in the response
        JSONArray movies = jsonResponse.getJSONArray("movies");

        // add each movie's title to an array
        String[] movieTitles = new String[movies.length()];
        for (int i = 0; i < movies.length(); i++)
        {
            JSONObject movie = movies.getJSONObject(i);
            movieTitles[i] = movie.getString("title");
            JSONArray cast = movie.getJSONArray("abridged_cast");
            String[] castmembers = new String[cast.length()];
            for(int j =0; j<cast.length();j++){


            }
        }

        // update the UI
        refreshMoviesList(movieTitles);
    }
    catch (JSONException e)
    {
        Log.d("Test", "Failed to parse the JSON response!");
    }
}

Answer 1

如果我理解你的问题，你希望你的输出格式如下：Matrix：Keane Reeves，Lawrence fishburne，.....
假设您从json获取正确的数据，以下内容将执行：

   // add each movie's title to an array
    String[] movieTitles = new String[movies.length()];
    for (int i = 0; i < movies.length(); i++)
    {
        JSONObject movie = movies.getJSONObject(i);
        movieTitles[i] = movie.getString("title") + ":";
        JSONArray cast = movie.getJSONArray("abridged_cast");
        String[] castmembers = new String[cast.length()];
        for(int j =0; j<cast.length();j++){
　　　　　　　　　　　　　　　　movieTitles[i] += castmembers[j];
                if( j < (cast.length()-1)){
                     movieTitles[i] += ",";
                 } 
        }
    }

如何从Rotten Tomatoes的API访问特定的JSON数据

1 个答案: