android - json 没有到达远程服务器

Question

我试图在远程服务器上运行一个应用程序，当我尝试注册一个用户时它返回了这个。 同一个应用程序在我的笔记本电脑上建立服务器时运行良好。我想知道我是否缺少对 LINUX 的任何依赖以在我的服务器端接收 JSON。

org.json.JSONException: End of input at character 0 of

我的 Android Activity 如下:-

      private void registerUser(final String reg_FullName_value ,final String  reg_Username_value, final String reg_Email_value, final String  reg_Password_value)
    {
        // Tag used to cancel the request
        String tag_string_req = "register";
        pDialog.setMessage("Registering ...");
        showDialog();
        StringRequest sr = new StringRequest(Request.Method.POST, AppURLs.URL , new Response.Listener<String>(){
            @Override
            public void onResponse(String response)
            {
                Log.d("Response",response);
                hideDialog();
                try {
                    JSONObject jObj = new JSONObject(response);
                    boolean error = jObj.getBoolean("error");
                    if (!error)
                    {
                        Toast.makeText(getApplicationContext(), "Regestered successfully", Toast.LENGTH_LONG).show();
                        Intent intent = new Intent(Register.this, Login.class);
                        startActivity(intent);
                        finish();
                        setContentView(R.layout.login);
                    }
                    else
                    {
                        Toast.makeText(getApplicationContext(), "Not Regestered.", Toast.LENGTH_LONG).show();
                    }
               }catch (JSONException e)
                {
                    e.printStackTrace();
                }
            }
        }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {
                Toast.makeText(getApplicationContext(),
                       "Some problem occur: check NetworK" , Toast.LENGTH_LONG).show();
                hideDialog();
            }
        }) {
            @Override
            protected Map<String, String> getParams()
            {
                Map<String, String> params = new HashMap<String, String>();
                params.put("tag", "register");
                params.put("FullName", reg_FullName_value );
                params.put("Username", reg_Username_value);
                params.put("Email", reg_Email_value);
                params.put("Password", reg_Password_value);
                return params;


}

        };

        AppController.getInstance().addToRequestQueue(sr, tag_string_req);
    }    

处理这个响应的 PHP 文件是这样的:-

<?php
error_reporting(E_ALL ^ E_DEPRECATED);
if(isset($_POST['tag']) && $_POST['tag'] != '')
{
  $tag = $_POST['tag'];
  require_once 'DB_Functions.php';
  $db = new DB_Functions();
  $response = array("tag" => $tag, "error" => FALSE);
  //registration
  if ($tag == 'register')
  {
    $name = $_POST['FullName'];
    $username = $_POST['Username'];
    $email = $_POST['Email'];
    $password = $_POST['Password'];
    $user = $db->storeUser($name, $username , $email , $password);
    if ($user)
    {
      $response["error"] = FALSE;
      $response["error_msg"] = "Registration Successful";
      echo json_encode($response);
    }
    else
    {
      $response["error"] = TRUE;
      $response["error_msg"] = "Registration Failed";
      echo json_encode($response);
    }
  }
  //login
  else if ($tag == 'login')
  {
    $user = $_POST['Username'];
    $password = $_POST['Password'];
    $user_check = $db->getUserByUserAndPassword($user , $password);
    if ($user_check != false)
    {
      $response["Success"] = TRUE;
      $response["error"] = FALSE;
      $response["error_msg"] = "User Exists";
      $response["Username"] = $user;
      echo json_encode($response);
    }
    else
    {
      $response["error"] = TRUE;
      $response["error_msg"] = "No such User";
      echo json_encode($response);
    }
  }
  else
  {
    {
      $response["error"] = TRUE;
      $response["error_msg"] = "Incorrect token or password!";
      echo json_encode($response);
    }
  }
}
else
{
    $response["error"] = TRUE;
    $response["error_msg"] = "Recieved No Data";
    echo json_encode($response);
}
?>

我正在使用 Droplet 制作云服务器。我需要知道背后的原因是什么。

Answer 1

试试这个。

if(TextUtils.isEmpty(response)){
     Toast.makeText(this, "response is null", Toast.LENGTH_SHORT).show();
     return;
}
Log.d("Response",response);

您应该首先检查 response 是否为 null 且不是空字符串。

如果您的响应为空，您的 JSONObject jObj = new JSONObject(response); 将得到一个异常。