php - 如何知道什么值从我的 Activity 发送到 Android 中的 MySql 服务器

Question

我正在使用 Volley 将数据发送到 MySQL。我不知道为什么我的数据没有插入 MySQL，我想查看从 Activity 发送到 MySQL 服务器的数据。下面是我的代码

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);
    }

我正在使用 PHP 将数据插入 MySQL。

我的 PHP 代码

 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

在 PHP 代码中只需删除这一行

$array[] = $response;

在json_encode($response);中传递$response