应用程序无法在PHP文件Android Studio中访问PHP文件或错误

Question

我的应用尝试在服务器中创建一个新行。 我得到的错误是jObj = new JSONObject(json); Null Point Exception jObj = new JSONObject(json);

这是创建新行的php文件：

<?php
$response = array();
if (isset($_POST['user']) && isset($_POST['pass']) &  isset($_POST['mail'])&&     isset($_POST['num']))
{
$user = $_POST['user'];
$pass = $_POST['pass'];
$mail = $_POST['mail'];
$num = $_POST['num'];
require_once __DIR__ . '/db_connect2.php';
$db = new DB_CONNECT();
$result = $db->query("INSERT INTO users(Name, Password, Email,ConfirmNum) VALUES('$user', '$pass', '$mail', '$num')");
if ($result) {
$response["success"] = 1;
$response["message"] = "user successfully created.";
echo json_encode($response);
} else {
$response["success"] = 0;
$response["message"] = "Oops! An error occurred.";
echo json_encode($response);
}
} else {
$response["success"] = 0;
$response["message"] = "missing fields";
echo json_encode($response);
} 
?>

请求更新的解析器是：

package com.example.denis.onthego;
import android.content.ContentValues;
import android.util.Log;

import org.json.JSONException;
import org.json.JSONObject;

import java.io.IOException;

import okhttp3.MediaType;
import okhttp3.OkHttpClient;
import okhttp3.Request;
import okhttp3.RequestBody;
import okhttp3.Response;


public class JSONParser {

static JSONObject jObj;
static String json;

// constructor
public JSONParser() {
}

// function get json from url
// by making HTTP POST or GET mehtod
public static JSONObject makeHttpRequest(String url, String method, ContentValues params) {
    // Making HTTP request
    try {
        final OkHttpClient client = new OkHttpClient();
        Request request;
        // check for request method
        if (method.equals("POST")) {
            // request method is POST

            MediaType contentType = MediaType.parse("application/x-www-form-urlencoded; charset=UTF-8");
            String content = "";
            for (String key : params.keySet())
            {
                if ( !content.isEmpty())
                    content += "&";

                content += key + "=" + params.get(key);
            }

            RequestBody body = RequestBody.create(contentType, content);
            request = new Request.Builder().url(url).post(body).build();
        }
        else  {
            // request method is GET
            request = new Request.Builder().url(url).build();
        }
        final Response response = client.newCall(request).execute();
        json = response.body().string();

    } catch (IOException e) {
        e.printStackTrace();
    }
    // 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;
}
}

对解析器的调用是：

JSONObject json = jsonParser.makeHttpRequest(url_create_product,"POST", params);

虽然“add_user”是用于创建新行的php文件（它是正确的URL），而“params”不是空的并且包含正确的键和内容。 解析器或php文件有问题吗？ 这是一个非常严肃的学校项目，这是我唯一缺少的。

以下是参数：

 params.put("user", "swane15");
 params.put("pass", "asdeg124A");
 params.put("mail", "asf@asd.com");
 params.put("num", "111111");

为什么php文件没有返回任何内容？ 是因为应用程序无法访问它或php文件本身是否有错误？

Answer 1

 MediaType JSON = MediaType.parse("Content-Type:application/json; charset=UTF-8");
 RequestBody body = RequestBody.create(JSON, params.toString());

改成：

      MediaType contentType = MediaType.parse("application/x-www-form-urlencoded; charset=UTF-8");

      String content = "";        
      for (String key : params.keySet())
      {
          if ( !content.isEmpty())
              content += "&";

          content += key + "=" + params.get(key);
      }

      RequestBody body = RequestBody.create(contentType, content);

要做得对，你应该URLEncoder.encode（）的值。