Question

我有＆＃34; POST＆＃34;的以下代码请求在哪里，我有一个API和参数。但是，在这里我没有得到任何回应也没有错误。请帮助我。

public JSONObject makeHttpRequest(String url, List<NameValuePair> params)       throws Exception {      
        // Making HTTP request
        try {
            Log.d("URL :", url);
            Log.d("params :", params.toString());
                // defaultHttpClient
                DefaultHttpClient httpClient = new DefaultHttpClient();
                HttpPost httpPost = new HttpPost(url);
                httpPost.setEntity(new UrlEncodedFormEntity(params));
                HttpResponse httpResponse = httpClient.execute(httpPost);
                HttpEntity httpEntity = httpResponse.getEntity();
                is = httpEntity.getContent();
           Log.d("InputStreem :", is.toString());
        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        } catch (ClientProtocolException e) {
          e.printStackTrace();
        } catch (IOException e) {
          e.printStackTrace();
        }
        try {
            BufferedReader reader = new BufferedReader(new InputStreamReader(is, "iso-8859-1"), 8);
            Log.d("BufferedReader :", reader.toString());
                  StringBuilder sb = new StringBuilder();
                  String line = null;
                  while ((line = reader.readLine()) != null) {
                    sb.append(line + "\n");
                  }
                  is.close();
                  jsonResponse = sb.toString();
            System.out.println("Respose :"+jsonResponse);//Here Resonse: "nothing"
        } catch (Exception e) {
            throw new Exception(e.getMessage());
        }
        // try parse the string to a JSON object
        try {
            jsonData = new JSONObject(jsonResponse);
        } catch (JSONException e) {
            throw new Exception(e.getMessage());
        }
        // return JSON String
        return jsonData;
    }

Answer 1

试试这个：

定义JSON-Class。例如：

`

 public class JSONParser {

        static InputStream is = null;
        static JSONObject jObj = null;
        static String json = "";

        // constructor
        public JSONParser() {

        }

        public JSONObject getJSONFromUrl(String url, List<NameValuePair> params) {

            // Making HTTP request
            try {
                // defaultHttpClient
                DefaultHttpClient httpClient = new DefaultHttpClient();
                HttpPost httpPost = new HttpPost(url);
                httpPost.setEntity(new UrlEncodedFormEntity(params));

                HttpResponse httpResponse = httpClient.execute(httpPost);
                HttpEntity httpEntity = httpResponse.getEntity();
                is = httpEntity.getContent();

            } catch (UnsupportedEncodingException e) {
                e.printStackTrace();
            } catch (ClientProtocolException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }

            try {
                BufferedReader reader = new BufferedReader(new InputStreamReader(
                        is, "iso-8859-1"), 8);
                StringBuilder sb = new StringBuilder();
                String line = null;
                while ((line = reader.readLine()) != null) {
                    sb.append(line + "\n");
                }
                is.close();
                json = sb.toString();
                Log.e("JSON", json);
            } catch (Exception e) {
                Log.e("Buffer Error", "Error converting result " + e.toString());
            }

            // try parse the string to a JSON object
            try {
                jObj = new JSONObject(json);
            } catch (JSONException e) {
                Log.e("JSON Parser", "Error parsing data " + e.toString());
            }

            // return JSON String
            return jObj;

        } }

定义Functions-Class，其中包含您的应用程序与服务器通信所需的所有功能。例如：

public class UserFunctions {

private JSONParser jsonParser;

// URL of the API
private static String functionURL = "http://xxx.xxx.xxx.xx/.../";

// TAG to determine which function/query should be executed on serverside
// e.g. with IF-ELSE
private static String function_tag = "getData";

// Constructor
public UserFunctions() {
    jsonParser = new JSONParser();
}

/**
 * FUNCTION GetSomeData
 **/

// Here you can define functions, and give parameters (if you need)
// and catch those on serverside with POST as their names
// you defined in qoutes, in this example you could catch the 2 parameters
// with
// $email = $_POST['email']; and $password = $_POST['password']; (assumed
// you use PHP)

public JSONObject functionGetData(String parameter1, String parameter2) {
    // Building Parameters
    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("tag", function_tag));
    params.add(new BasicNameValuePair("email", parameter1));
    params.add(new BasicNameValuePair("password", parameter1));
    JSONObject json = jsonParser.getJSONFromUrl(functionURL, params);
    return json;
}}

使用AsynchTask获取json数据。例如：

私有类FetchDataTask扩展 AsyncTask {

    String parameter1, parameter2;

    @Override
    protected void onPreExecute() {
        super.onPreExecute();
        // Define variables here (if needed), that will be commited to
        // function/method in
        // doInBackground() as parameters

    }

    @Override
    protected JSONObject doInBackground(String... args) {

        UserFunctions userFunction = new UserFunctions();

        JSONObject json = userFunction.functionGetData(parameter1,
                parameter2);

        return json;

    }

    @Override
    protected void onPostExecute(JSONObject json) {

        try {

            // Catch data from JSON
            // Create JSON object and get it by calling method
            // getString("data")
            // on serverside json:
            // $response["jsonObj"]["data"] = $SQL_result['columnXY'];

            JSONObject json_object = json.getJSONObject("jsonObj");

            String data = json_object.getString("data");

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

        }

    }}

希望它有所帮助。

没有获得JSON响应

1 个答案: