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- http://codeforces.com/api/user.info?handles=DmitriyH;

中提取数据时

但我只想要用户的“firstName”。任何人都可以推荐更改我的代码吗？

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;

}
}

修改

现在假设我正在使用网址＆＃34; codeforces.com/api/user.status？handle = avm12＆＃34 ;;我想提取前十个（或者说n个数字）＆＃34;问题＆＃34;标签。那我该怎么办？

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添加到您的网址，以将输出限制为所需的N个结果......

Answer 2

您的完整JSONObject包含一个标记为"result"的数组，因此您应该像这样提取firstName：

JSONArray result = json.getJSONArray("result");
JSONObject jo = result.getJSONObject(0);
String firstName = jo.getString("firstName");

Answer 3

您的第一个代码是JSONArray，因此您需要制作一个返回JSONArray而不是JSONObject的方法。根据您的要求，您可以通过

获得它

 jArray.getJSONObject(0).get("firstName");

Answer 4

在您发布的代码中未使用正确的网址。以下作品：

final static String URL = http://codeforces.com/api/user.info?handles=;

...

protected String doInBackground(String... arg0) {
        // TODO Auto-generated method stub
        try {
            json = lastSub("DmitriyH");
            final JSONArray result = json.getJSONArray("result");
            final JSONObject jsonObject = result.getJSONObject(0);
            return jsonObject.getString("firstName");
        } catch (ClientProtocolException e) {
            // TODO Auto-generated catch block
....

JSONObject返回类型

4 个答案: