java - JSONObject 返回类型

Question

Java 中的以下 android 应用程序代码输出到:

[{"handle":"DmitriyH","firstName":"Dmitriy","lastName":"Khodyrev","country":"Russia","city":"Moscow","organization":"KL","contribution":122,"rank":"master","rating":2040,"maxRank":"international master","maxRating":2072,"lastOnlineTimeSeconds":1432130513,"registrationTimeSeconds":1268570311}]}

从 codeforces api 中提取数据时- http://codeforces.com/api/user.info?handles=DmitriyH ;

但我只想要用户的“firstName”。 任何人都可以建议更改我的代码吗？？

public class Http extends Activity {
   TextView httpStuff;
   HttpClient client;
   JSONObject json;
   final static String URL = http://codeforces.com/api/user.info?handles=;


@Override
protected void onCreate(Bundle savedInstanceState) {
    // TODO Auto-generated method stub
    super.onCreate(savedInstanceState);
    setContentView(R.layout.httpex);

    httpStuff = (TextView)findViewById(R.id.tvHttp);
    client = new DefaultHttpClient();
    new Read().execute("result");
}

public JSONObject lastSub(String username) throws ClientProtocolException, IOException, JSONException {
    StringBuilder url = new StringBuilder(URL);
    url.append(username);
    HttpGet get = new HttpGet(url.toString());

    HttpResponse r = client.execute(get);
    //httpStuff.setText("xxx");
    int status = r.getStatusLine().getStatusCode();
    if(status == 200) {
        HttpEntity e = r.getEntity();
        String data = EntityUtils.toString(e);
        JSONObject last = new JSONObject(data);

        return last;
    }
    else {
        Toast.makeText(Http.this, "error", Toast.LENGTH_SHORT);
        return null;

    }
}

public class Read extends AsyncTask <String, Integer, String> {

    @Override
    protected String doInBackground(String... arg0) {
        // TODO Auto-generated method stub
        try {
            json = lastSub("avm12");
            return json.getString(arg0[0]);
        } catch (ClientProtocolException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (JSONException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        return null;
    }

    @Override
    protected void onPostExecute(String result) {
        // TODO Auto-generated method stub
        httpStuff.setText(result);
    }

}

Answer 1

public JSONObject lastSub(String username) throws ClientProtocolException, IOException, JSONException {
StringBuilder url = new StringBuilder(URL);
url.append(username);
HttpGet get = new HttpGet(url.toString());

HttpResponse r = client.execute(get);
//httpStuff.setText("xxx");
int status = r.getStatusLine().getStatusCode();
if(status == 200) {
    HttpEntity e = r.getEntity();
    String data = EntityUtils.toString(e);
    JSONObject last = new JSONObject(data).getJSONArray("result").getJSONObject(0);

    return last;
}
else {
    Toast.makeText(Http.this, "error", Toast.LENGTH_SHORT);
    return null;

}
}

编辑

Now suppose I am using the url "codeforces.com/api/user.status?handle=avm12"; and I want to extract the first ten( or say n number of them) "problem" tags. What should I do then?

那么首先获取根 JSONArray

JSONArray array= new JSONObject(data).getJSONArray("result");

然后在 for 循环中获取所有需要的 JSONObjects

for(int k=0;k<n;k++){ // n must be less than array.length()
    JSONObject problemObject=array.getJSONObject(k).getJSONObject("problem");
   //do what you need to do with the problemObject E.G. Add them to an ArryList... 
}

考虑将 &count=N 添加到您的 url 以将输出限制为所需的 N 个结果...