使用mysql数据库在android中验证用户名和密码的登录页面

Question

我使用了这个链接https://www.simplifiedcoding.net/android-mysql-tutorial-android-login-using-php-mysql/ 登录页面。但是当我输入注册用户名和密码时，我无法进入下一个活动。这是我的代码：

 public static final String USER_NAME = "USER_NAME";
    public static final String PASSWORD = "PASSWORD";
    private static final String LOGIN_URL = "http://192.xxx.x.xx/VolleyUpload/login.php";
    private EditText editTextUserName;
    private EditText editTextPassword;
    private Button buttonLogin;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_activity_login);

        editTextUserName = (EditText) findViewById(R.id.username);
        editTextPassword = (EditText) findViewById(R.id.password);
        buttonLogin = (Button) findViewById(R.id.buttonUserLogin);
        buttonLogin.setOnClickListener(this);
    }

    private void login(){
        String username = editTextUserName.getText().toString().trim();
        String password = editTextPassword.getText().toString().trim();
        userLogin(username,password);
    }

    private void userLogin(final String username, final String password){
        class UserLoginClass extends AsyncTask<String,Void,String>{
            ProgressDialog loading;
            @Override
            protected void onPreExecute() {
                super.onPreExecute();
                loading = ProgressDialog.show(ActivityLogin.this,"Please Wait",null,true,true);
            }

            @Override
            protected void onPostExecute(String s) {
                super.onPostExecute(s);
                loading.dismiss();
                if(s.equalsIgnoreCase("success")){
                    Intent intent = new Intent(ActivityLogin.this,UserProfile.class);
                    intent.putExtra(USER_NAME,username);
                    startActivity(intent);
                }else{                   Toast.makeText(ActivityLogin.this,s,Toast.LENGTH_LONG).show();
                }
            }

            @Override
            protected String doInBackground(String... params) {
                HashMap<String,String> data = new HashMap<>();
                data.put("username",params[0]);
                data.put("password",params[1]);

                RegisterUserClass ruc = new RegisterUserClass();
                String result = ruc.sendPostRequest(LOGIN_URL,data);
                return result;
            }
        }
        UserLoginClass ulc = new UserLoginClass();
        ulc.execute(username,password);
    }

    @Override
    public void onClick(View v) {
        if(v == buttonLogin){
            login();
        }
    }

这是我的PHP代码：

<?php
 if($_SERVER['REQUEST_METHOD']=='POST'){
 $username = $_POST['username'];
 $password = $_POST['password'];

 require_once('dbConnect.php');

 $sql = "select * from users where username='$username' and password='$password'";

 $check = mysqli_fetch_array(mysqli_query($con,$sql));

 if(isset($check)){
 echo "success";
 }else{
 echo "Invalid Username or Password";
 }

 }else{
 echo "error try again";
 }
 ?>

谁能告诉我哪里做错了？在方法onPostExecute（String s）中，我没有收到“成功”消息。我想如果我得到了，我就可以进入下一个活动。