Question

我想使用POST方法传递2个参数，并从使用php开发的API获得该响应。我想使用GSON和Volley。

我已经编写了一个方法并在onCreate方法中调用它。这是我的方法类-

private void loadDetailed() {
        //Toast.makeText(DetailedBoatActivity.this, "Called", Toast.LENGTH_SHORT).show();
        Intent i = getIntent();
        Integer boat_id = i.getExtras().getInt("boat_id");
        Integer owner_id = i.getExtras().getInt("owner_id");
        Gson gson = new Gson();
        Map<String, Integer> jsonMap = new HashMap<>();
        jsonMap.put("boat_id", boat_id);
        jsonMap.put("owner_id", owner_id);
        JSONObject jsonObject = new JSONObject(jsonMap);
        Log.d("jsonObject", String.valueOf(jsonObject));
        JsonObjectRequest jsonObjectRequest = new JsonObjectRequest(Request.Method.POST, DETAILED_BOAT_URL, jsonObject, new Response.Listener<JSONObject>() {
            @Override
            public void onResponse(JSONObject response) {
                try {
                    Toast.makeText(DetailedBoatActivity.this, "Try", Toast.LENGTH_SHORT).show();
                    JSONArray array = new JSONArray(response);
                    Log.i("array1", String.valueOf(array));

                } catch (JSONException e) {
                    Toast.makeText(DetailedBoatActivity.this, "Error", Toast.LENGTH_SHORT).show();
                    e.printStackTrace();
                }
            }
        }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {

            }
        });
        Volley.newRequestQueue(this).add(jsonObjectRequest);
    }

但是我没有从我的API得到任何响应。这是我的代码-

public function boatDetails($id, $owner_id){
            $stmt = $this->con->prepare("SELECT * FROM `boat_details` WHERE id=".$id);
            $stmt->execute();
            $resultSet = $stmt->get_result();
            $boat_details = array();
            while($result = $resultSet->fetch_assoc()){
                $boat_details[] = $result;
            }

            return $boat_details;   
        }

我已手动检查。我通过使用以下代码获取JSON格式的值-

require_once '../includes/dboperations.php';

    $response = array();

    if($_SERVER['REQUEST_METHOD'] == 'POST'){

        //if($_GET['boat_id']){

            $db = new DbOperations();

            $resultSet = $db->boatDetails($_POST['boat_id'], $_POST['owner_id']);

            echo json_encode($resultSet);
        //}
    }

请帮助我传递参数，并返回对显示实体的响应。

Answer 1

    String url = URL.LOGIN_URL + "&mail=" + email + "&pass=" + pwd + ""; 
RequestQueue queue;
queue = Volley.newRequestQueue(YOUR_ACTIVITY_NAME.this);
StringRequest req = new StringRequest(Request.Method.GET, url, new Response.Listener<String>() {
            @Override
            public void onResponse(String response) {   }
        }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {

                System.out.println("error ");
                error.printStackTrace();
            }
        });
        queue.add(req);

如何使用post方法传递参数并从url获取响应并使用gson和volley在android中显示它？

1 个答案: