Question

过去4天我一直想弄清楚这一点，我准备把头发拉出来！

我需要你的帮助，任何人都可以告诉我我做错了什么吗？

以下是我用于开发的JSON格式链接：已删除隐私权问题

这是我的代码：

 public class JSONActivity extends Activity {


TextView http;
HttpClient client;
JSONObject json;


final static String URL = "REMOVED FOR PRIVACY CONCERNS

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    http = (TextView) findViewById(R.id.http);
    client = new DefaultHttpClient();
    new Read().execute("firstName");


}

public JSONObject getpw(String password)
        throws ClientProtocolException, IOException, JSONException {
    StringBuilder url = new StringBuilder(URL);
    url.append(password);

    HttpGet get = new HttpGet(url.toString());
    HttpResponse r = client.execute(get);
    int status = r.getStatusLine().getStatusCode();
    if (status == 200) {
        HttpEntity e = r.getEntity();
        String data = EntityUtils.toString(e);
        JSONArray getname = new JSONArray(data);
        JSONObject last = getname.getJSONObject(3);
        return last;
    } else {
        Toast.makeText(JSONActivity.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=getpw("trustme");
            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
        http.setText(result);
    }

}}

非常感谢任何帮助。

Answer 1

您期待JSONArray，但webservice只是返回一个JSONObject。

我做了以下更改：

JSONArray getname = new JSONArray(data); - ＆gt; JSONObject getname = new JSONObject(data);

在异步任务中，我做了以下更改：

return json.getString("firstName");

这是完整的代码：

public class PlaygroundActivity extends Activity {
    TextView http;
    HttpClient client;
    JSONObject json;

    final static String URL = "REMOVED FOR PRIVACY CONCERNS

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
        http = (TextView) findViewById(R.id.http);
        client = new DefaultHttpClient();
        new Read().execute("firstName");

    }

    public JSONObject getpw(String password) throws ClientProtocolException,
            IOException, JSONException {
        StringBuilder url = new StringBuilder(URL);
        url.append(password);

        HttpGet get = new HttpGet(url.toString());
        HttpResponse r = client.execute(get);
        int status = r.getStatusLine().getStatusCode();
        if (status == 200) {
            HttpEntity e = r.getEntity();
            String data = EntityUtils.toString(e);
            JSONObject getname = new JSONObject(data);

            return getname;
        } else {
            Toast.makeText(PlaygroundActivity.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 = getpw("trustme");
                return json.getString("firstName");
            } 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
            http.setText(result);
        }
    }
}

Answer 2

您可以使用以下代码从服务器获取响应，然后解析json。

public static void extractParameters(){
    String urlRequest = "REMOVED FOR PRIVACY CONCERNS";
    String response = getServerResponse(urlRequest);
    try {
        JSONObject json = new JSONObject(response);
        boolean suspended = json.getBoolean("suspended");
        String firstName = json.getString("firstName");
        String lastName = json.getString("lastName");
        int checkdin = json.getInt("checkedin");
        int checkindatetime = json.getInt("checkindatetime");
        //Get Json object address
        JSONObject address = json.getJSONObject("address");
        String streedAddress = address.getString("streetAddress");//In same way get city etc
        //Get PhoneNumber Array
        JSONArray phoneNumbers = json.getJSONArray("phoneNumber");
        String type = phoneNumbers.getJSONObject(0).getString("type");
        String number = phoneNumbers.getJSONObject(0).getString("number");// and so on for 1,2,3...


    } catch (Exception e) {
        // TODO: handle exception
    }

}

这些方法有助于从服务器获取响应

/**
 * 
 * @param urlRequest URL of server 
 * @return
 */
public static String getServerResponse(String urlRequest){
    Log.d("urlRequest",urlRequest);
    String response = "";
    HttpURLConnection conn = null;
    try {
        conn = (HttpURLConnection) new URL(urlRequest).openConnection();
    } catch (MalformedURLException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

    try {
        response = read(conn.getInputStream());
    } catch (IOException e) {
        e.printStackTrace();
    }

    Log.d("response",response);
    return response.trim();
}
private static String read(InputStream in) throws IOException {
    StringBuilder sb = new StringBuilder();
    BufferedReader r = new BufferedReader(new InputStreamReader(in), 1000);
    for (String line = r.readLine(); line != null; line = r.readLine()) {
        sb.append(line);
    }
    in.close();
    return sb.toString();
}

您发送的JSON响应示例为：

{
  "suspended": "false",
  "firstName": "John",
  "lastName": "Smith",
  "checkedin": 0,
  "checkindatetime": 0,
  "address": {
    "streetAddress": "21 2nd Street",
    "city": "New York",
    "state": "NY",
    "postalCode": "10021"
  },
  "phoneNumber": [
    {
      "type": "home",
      "number": "212 111-1111"
    },
    {
      "type": "fax",
      "number": "646 222-2222"
    }
  ]
}

无法解析JSON在我的应用程序中

2 个答案: