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如何使用Java解析android中从facebook检索到的JSONObject?_Java_Android_Json_Facebook_Facebook Graph Api - Fatal编程技术网

如何使用Java解析android中从facebook检索到的JSONObject?

如何使用Java解析android中从facebook检索到的JSONObject?,java,android,json,facebook,facebook-graph-api,Java,Android,Json,Facebook,Facebook Graph Api,我试图用java解析从facebook图形API返回的JSONObject。 我想通过graph api获取一些用户数据和朋友的数据。 我试图分析的图形如下: { "id": "985453561536987", "first_name": "abc", "last_name": "xyz", "name": "abc xyz", "birthday": "03/17/1995", "gender": "male", "friends": { "data":

我试图用java解析从facebook图形API返回的JSONObject。 我想通过graph api获取一些用户数据和朋友的数据。 我试图分析的图形如下:

{
  "id": "985453561536987",
  "first_name": "abc",
  "last_name": "xyz",
  "name": "abc xyz",
  "birthday": "03/17/1995",
  "gender": "male",
  "friends": {
    "data": [
      {
        "first_name": "Div",
        "last_name": "Tol",
        "gender": "female",
        "id": "150534786284567"
      }
    ],
    "paging": {
      "next": "https://graph.facebook.com/v2.4/985363151536997/friends?fields=first_name,last_name,about,birthday,gender&access_token=CAANEU8qO1HYBAJG4ZCYtc5lLuZBv3zKWe6qRN3Oy5iYtijFsQwplDH2AFe2YiXS1lZCXtxnYZBXLhBo6N65KEmW3wuNKxXiZCCVgFpg8r3elkqgCRFZBZAIgscoZCtxezk08fpiPnYqujCdMY4ZB3d88vqcDZCnpon9TjPZBE2Irzlrcqn8gzPZB8nvRu2ZBoLt6Y1yGlovgt0K18uYBbyj8GpNZCX&limit=25&offset=25&__after_id=enc_AdBfWxVZCaZAlL5ZBGlZAQjMZC2M5m4uQMNSQPfwNfsxC9TwgQzgK55kQEVGeZBEJMoyXfvySnZAOwaHCGvmgj5Ngzut6D9"
    },
    "summary": {
      "total_count": 533
    }
  },
}
我在尝试解析JSONObject时遇到NullPointerException 对于解析上述信息,我执行以下操作:

public void graphActivity(AccessToken accessToken) {

        //used is graph request code from graph api explorer
        GraphRequest request = GraphRequest.newMeRequest(
                accessToken,
                new GraphRequest.GraphJSONObjectCallback() {
                    @Override
                    public void onCompleted(JSONObject object, GraphResponse response) {
                        // Insert your code here
                        try {

                            //getting error in following code where object.getString("first_name");
                            //is written
                            //null pointer exception

                            //userVO.setId(object.getString("id");
                            userVO.setUserFirstName(object.getString("first_name"));
                            userVO.setUserLastName(object.getString("last_name"));
                            userVO.setUserName(object.getString("name"));
                            userVO.setUserBirthday(object.getString("birthday"));
                            userVO.setUserGender(object.getString("gender"));


                            JSONObject friendsJsonObject = object.getJSONObject("friends");

                            JSONArray data = friendsJsonObject.getJSONArray("data");

                            String userString = "USER::\n"
                                    + "id: " + userVO.getId() + "\n"
                                    + "fn: " + userVO.getUserFirstName() + "\n"
                                    + "ln: " + userVO.getUserLastName() + "\n"
                                    + "un: " + userVO.getUserName() + "\n"
                                    + "gen: " + userVO.getUserGender() + "\n"
                                    + "bday: " + userVO.getUserBirthday() + "\n";

                            Log.d(TAG, userString);
                            friendList = new ArrayList();
                            int length_list = data.length();
                            for (int i = 0; i < length_list; i++) {
                                JSONObject dataObj = data.getJSONObject(i);
                                //friendVO.setFriendId(dataObj.getString("id"));

                                friendVO.setFriendFirstName(dataObj.getString("first_name"));
                                friendVO.setFriendLastName(dataObj.getString("last_name"));
                                friendVO.setFriendName(dataObj.getString("name"));
                                friendVO.setFriendBirthday(dataObj.getString("birthday"));
                                friendVO.setFriendGender(dataObj.getString("gender"));
                                friendList.add(friendVO);

                                String friendString = "FRIEND::\n"
                                        + "id: " + friendVO.getFriendId() + "\n"
                                        + "fn: " + friendVO.getFriendFirstName() + "\n"
                                        + "ln: " + friendVO.getFriendLastName() + "\n"
                                        + "un: " + friendVO.getFriendName() + "\n"
                                        + "gen: " + friendVO.getFriendGender() + "\n"
                                        + "bday: " + friendVO.getFriendBirthday() + "\n";

                                Log.d(TAG, friendString);




                            }


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

                    }
                });

        Bundle parameters = new Bundle();
        parameters.putString("fields", "id,first_name,last_name,name,birthday,about,gender,website,friends{first_name,last_name,name,about,birthday,gender}");
        request.setParameters(parameters);
        request.executeAsync();
    }

您可以使用com.google.gson.gson类中的fromJson方法将json转换为java对象,而不是使用所有setter。教程。
这将简化代码并可能解决问题。

您可以使用com.google.gson.gson类中的fromJson方法将json转换为java对象,而不是使用所有setter。教程。
这将简化您的代码并可能解决问题。

基本上,GraphResponse在userVO.setUserFirstNameobject.getStringfirst\u name行之前包含您期望的响应JSONObject;请添加以下行并检查


基本上,GraphResponse在userVO.setUserFirstNameobject.getStringfirst\u name行之前包含您期望的响应JSONObject;请添加以下行并检查


请尝试下面的方法。我在我的一个项目中使用了同样的方法,效果很好

public Map<String, String> getGraphData(String fbGraph) {
        Gson gson = new Gson();//.toJson(fbGraph);
        Map<String, String> fbProfile = gson.fromJson(fbGraph, new TypeToken<Map<String, String>>(){}.getType());//fromJson(gson, new TypeToken<Map<String, String>>(){}.getType());
        return fbProfile;
    }

请尝试下面的方法。我在我的一个项目中使用了同样的方法,效果很好

public Map<String, String> getGraphData(String fbGraph) {
        Gson gson = new Gson();//.toJson(fbGraph);
        Map<String, String> fbProfile = gson.fromJson(fbGraph, new TypeToken<Map<String, String>>(){}.getType());//fromJson(gson, new TypeToken<Map<String, String>>(){}.getType());
        return fbProfile;
    }
请尝试以下代码:

GraphRequest request = GraphRequest.newMeRequest(
            loginResult.getAccessToken(),
            new GraphRequest.GraphJSONObjectCallback() {
            @Override
            public void onCompleted(
                 final JSONObject object,
                 GraphResponse response) {
                 // Application code
                 final JSONObject jsonObject = response.getJSONObject();
                 String first_name = "";
                 String last_name = "";
                 try {
                      first_name = jsonObject.getString("first_name");
                      last_name =  jsonObject.getString("last_name"); 
                      JSONObject friends = jsonObject.getJSONObject("friends");
                      JSONArray data = friends.getJSONArray("data");
                      JSONObject objectdata = data.getJSONObject(0);
                      String friend_first_name = objectdata.getString("first_name");
                      String friend_last_name = objectdata.getString("last_name");
                 } catch (JSONException e) {
                      e.printStackTrace();
                 }});
请尝试以下代码:

GraphRequest request = GraphRequest.newMeRequest(
            loginResult.getAccessToken(),
            new GraphRequest.GraphJSONObjectCallback() {
            @Override
            public void onCompleted(
                 final JSONObject object,
                 GraphResponse response) {
                 // Application code
                 final JSONObject jsonObject = response.getJSONObject();
                 String first_name = "";
                 String last_name = "";
                 try {
                      first_name = jsonObject.getString("first_name");
                      last_name =  jsonObject.getString("last_name"); 
                      JSONObject friends = jsonObject.getJSONObject("friends");
                      JSONArray data = friends.getJSONArray("data");
                      JSONObject objectdata = data.getJSONObject(0);
                      String friend_first_name = objectdata.getString("first_name");
                      String friend_last_name = objectdata.getString("last_name");
                 } catch (JSONException e) {
                      e.printStackTrace();
                 }});
java.lang.NullPointerException发生在object.getStringfirst\u name;因为GraphRequest返回的JSONObject为null

当您使用request.executeAsync异步调用GraphRequest时;将在单独的线程中异步执行,即此行下面的代码未被阻止

因此,为了避免NPE问题,您可以将您的代码包含在以下内容中:

if(object!=null) { 
  //your code 
} 
else {
  Toast.makeText(this, "Please Try Again ",Toast.LENGTH_SHORT).show();
}
要解析JSON数据,请使用以下JSON解析器:

GSON- 杰克逊- java.lang.NullPointerException发生在object.getStringfirst\u name;因为GraphRequest返回的JSONObject为null

当您使用request.executeAsync异步调用GraphRequest时;将在单独的线程中异步执行,即此行下面的代码未被阻止

因此,为了避免NPE问题,您可以将您的代码包含在以下内容中:

if(object!=null) { 
  //your code 
} 
else {
  Toast.makeText(this, "Please Try Again ",Toast.LENGTH_SHORT).show();
}
要解析JSON数据,请使用以下JSON解析器:

GSON- 杰克逊-
你能发布错误日志吗?@muilpp是的,我用日志编辑了我的问题。你能发布错误日志吗?@muilpp是的,我用log编辑了我的问题。所以我需要使用graph response并将其转换为responseJSONObject而不是JSONObject??我引用文档以便我的感受……或者,您可以使用response Object检查错误,所以我需要使用graph response并将其转换为responseJSONObject??我引用文档以便我的感觉……或者,您可以使用Response对象检查错误