Wie kann ich wissen, welche Werte von meiner Aktivität an MySql Server in Android gesendet werden

Question

Ich benutze Volley, um Daten an MySQL zu senden. Ich weiß nicht, warum meine Daten nicht in MySQL eingefügt werden, ich möchte die Daten sehen, die von Activity zu MySQL Server gesendet werden. unten ist mein Code

private void SendDatatoserver(final String name,final String bname,final String location,final String phone, 
       final String website,final String disc,final String ownerphone,final String gname,final String latlong, final String spinner) 
    { 

     String tag_string_req = "req_vend_reg"; 

     pDialog.setMessage("Registering Please Wait ..."); 
     showDialog(); 

     StringRequest strReq = new StringRequest(Request.Method.POST, 
       AppURLs.Vend_URL, new Response.Listener<String>() { 

      @Override 
      public void onResponse(String response) { 
       hideDialog(); 
       //Log.d(TAG, response.toString()); 

       try { 
        JSONObject jObj = new JSONObject(response); 
        boolean error = jObj.getBoolean("error"); 

        if (!error) { 
         Intent intent = new Intent(AddNewBusiness.this, SuccessBsub.class); 
         startActivity(intent); 
         finish(); 
        } else { 
         String errorMsg = jObj.getString("error_msg"); 
         Toast.makeText(getApplicationContext(), 
           errorMsg, Toast.LENGTH_LONG).show(); 
        } 
       } catch (JSONException e) { 
        e.printStackTrace(); 
       } 

      } 
     },new Response.ErrorListener() { 

      @Override 
      public void onErrorResponse(VolleyError error) { 
       Toast.makeText(getApplicationContext(), 
         error.getMessage(), Toast.LENGTH_LONG).show(); 
       hideDialog(); 
      } 
     }){ 

      @Override 
      protected Map<String, String> getParams() { 
       // Posting params to register url 
       Map<String, String> params = new HashMap<String, String>(); 
       params.put("tag", "vregister"); 
       params.put("name", name); 
       params.put("bname", bname); 
       params.put("address", location); 
       params.put("phoneno", phone); 
       params.put("website", website); 
       params.put("disc", disc); 
       params.put("ownerphone", ownerphone); 
       params.put("gname", gname); 
       params.put("latlong", latlong); 
       params.put("spinner", spinner); 
       return params; 
      } 
     }; 

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

i PHP bin mit den Daten in MySQL einzufügen.

meine PHP-Code

if($tag == 'vregister'){ 

      $vname =$_POST['name']; 
      $bname =$_POST['bname']; 
      $vaddress =$_POST['address']; 
      $vphoneno =$_POST['phoneno']; 
      $vwebsite =$_POST['website']; 
      $vdisc =$_POST['disc']; 
      $vownerphone =$_POST['ownerphone']; 
      $vgname =$_POST['gname']; 
      $vlatlong =$_POST['latlong']; 
      $vspinner =$_POST['spinner']; 

      $status='Pending'; 


      //insert and get response 
      $sql=$dbh->prepare("INSERT INTO `newvenderreq`(pname, bname, address, phone, website, ownerphone, disc, gname, latlng, date, status, spinner) VALUES (:pname, :bname, :address, :phone, :website, :ownerphone, :disc, :gname, :latlng, NOW(), :status, :spinner)"); 
      $sql->execute(array(':pname'=>$vname, 
           ':bname'=>$bname, 
           ':address'=>$vaddress, 
           ':phone'=>$vphoneno, 
           ':website'=>$vwebsite, 
           ':ownerphone'=>$vownerphone, 
           ':disc'=>$vdisc, 
           ':gname'=>$vgname, 
           ':latlng'=>$vlatlong, 
           ':status'=>$status, 
           ':spinner'=>$vspinner)); 

      if ($sql->rowCount() > 0) { 
       // user stored successfully 
       $response["error"] = FALSE; 
       //$response["uid"] = $user["id"]; 
       //$response["bname"]= $bname; 

       header('Content-Type:Application/json'); 
       $array[] = $response; 
       echo json_encode($array); 
      } else { 
       // user failed to store 
       $response["error"] = TRUE; 
       $response["error_msg"] = "Error occured in Registartion"; 
       echo json_encode($response); 
      } 

     } 

Answer 1

In PHP-Code entfernen Sie diese Zeile

$array[] = $response; 

Pass $ response in json_encode($response);