6

我正在开发一个与 facebook 集成的 android 应用程序。我正在使用 fql 查询从 facebook 获取信息。我的fql方法是

                void runfql(){
                String fqlQuery = "SELECT uid, name, pic_small,birthday FROM user WHERE uid IN " +
                  "(SELECT uid2 FROM friend WHERE uid1 = me() )";
                Bundle params = new Bundle();
                params.putString("q", fqlQuery);
                Session session = Session.getActiveSession();
                Request request = new Request(session,
                "/fql",                         
                params,                         
                HttpMethod.GET,                 
                new Request.Callback(){         
                        public void onCompleted(Response response) {

                        JSONObject myjson=response.getGraphObject().getInnerJSONObject();
                        Log.d("ResultResultResult: " ,""+myjson);                           
                        }                  
                    }); 
                    Request.executeBatchAsync(request);  

                }

myjson 对象包含我想要的所有信息。像这样

 {"data":[{"uid":536089174,"birthday":"July 22","name":"Usman Aslam Sheikh"},{"uid":581379174,"birthday":"July 26","name":"Ammar Khalid"}

问题是如何将该信息存储到不同的数组中？

请为此目的纠正一些代码。？

4

3 回答 3

10 
String jsonString = yourstring;
JSONObject jsonResult = new JSONObject(jsonString);
JSONArray data = jsonResult.getJSONArray("data");
if(data != null) {
    String[] names = new String[data.length()];
    String[] birthdays = new String[data.length()];
    for(int i = 0 ; i < data.length() ; i++) {
        birthdays[i] = data.getString("birthday");
        names[i] = data.getString("name");
    }
}

检查http://www.androidhive.info/2012/01/android-json-parsing-tutorial/

于 2013-04-10T06:53:10.217 回答
0

像这样编辑您的代码...这可能会对您有所帮助..

JSONObject resultObject = new JSONObject(response);

JSONArray JArray = resultObject.getJSONArray("data");

                for (int t=0; t<JArray.length(); t++) {

                    JSONObject JObject = JtArray.getJSONObject(t);



builder.append(JObject.getString("uid")+": ");
于 2013-04-10T07:00:12.497 回答
0

在java中，它被解决为

String jsonString = yourstring;
JSONObject data = JSON.parse(jsonString);

//reading data using index
long firstData_uid = data["data"][0]["uid"];
String firstData_birthday = data["data"][0]["birthday"];
String firstData_name = data["data"][0]["name"];

在javascript中，它被解决为

var jsonString = yourstring;
var data = JSON.parse(jsonString);

//reading data using index
var firstData_uid = data["data"][0]["uid"];
var firstData_birthday = data["data"][0]["birthday"];
var firstData_name = data["data"][0]["name"];
于 2018-07-05T03:41:29.100 回答