Android 未连接到数据库以检查用户名和密码

我在教程视频中看到了这段代码，在该视频中，该应用程序运行良好，甚至显示“登录成功”吐司。

但是当我尝试复制它时，即使我在数据库中输入了正确的用户名和密码，它也对我不起作用。（我尝试使用用户的 LAN IP 和 Wi-Fi IP 使用 XAMPP，还尝试在 000webhost.com 上托管 PHP 脚本，但仍然无法使用任何一种方法）

我的安卓代码：

package com.example.devarsh.googlemapsbustracker;


import android.app.Dialog;
import android.content.Intent;
import android.os.Bundle;
import android.support.v7.app.AppCompatActivity;
import android.util.Log;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.Toast;

import com.android.volley.AuthFailureError;
import com.android.volley.Request;
import com.android.volley.RequestQueue;
import com.android.volley.Response;
import com.android.volley.VolleyError;
import com.android.volley.toolbox.StringRequest;
import com.android.volley.toolbox.Volley;
import com.google.android.gms.common.ConnectionResult;
import com.google.android.gms.common.GoogleApiAvailability;

import java.util.HashMap;
import java.util.Map;

public class MainActivity extends AppCompatActivity {

    private static final String TAG = "MainActivity";
    private static final int ERROR_DIALOG_REQUEST = 9001;
    private EditText User_name, Pass_word;
    private Button btnlogin;
    private static String URL_LOGIN = 
    "http://192.168.50.127/Android/login.php";


    //vars
    EditText enroll;


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

        btnlogin = findViewById(R.id.button2);
        User_name = findViewById(R.id.username);
        Pass_word = findViewById(R.id.password);

        btnlogin.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View view) {
                Login();
            }
        });


        isServicesOK();
    }

    private void Login(){

        String URL = 
    "imagebustracker.000webhostapp.com/Android/parul/login.php";
        RequestQueue requestQueue = Volley.newRequestQueue(this);
        StringRequest stringRequest = new StringRequest(Request.Method.POST, URL, new Response.Listener<String>() {
            @Override
            public void onResponse(String response) {

                if(response.trim().equals("success")){
                    Toast.makeText(getApplicationContext(), "Login Successfull", Toast.LENGTH_LONG).show();
                    Intent intent = new Intent(getApplicationContext(), Selection_Activity.class);
                    startActivity(intent);
                }else{
                    Toast.makeText(getApplicationContext(), "Login Unsuccessfull", Toast.LENGTH_LONG).show();
                }

            }
        }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {
                Toast.makeText(getApplicationContext(), "Login Unsuccessfull", Toast.LENGTH_LONG).show();
            }
        }) {
            @Override
            protected Map<String, String> getParams() throws AuthFailureError {

                Map<String, String> params = new HashMap<>();
                params.put("UserName", User_name.getText().toString().trim());
                params.put("PassWord", Pass_word.getText().toString().trim());

                return params;
            }
        };

        requestQueue.add(stringRequest);

    }


    public void isServicesOK() {
        Log.d(TAG, "isServicesOK: checking google services version");

        int available = GoogleApiAvailability.getInstance().isGooglePlayServicesAvailable(this);

        if(available == ConnectionResult.SUCCESS){
            //everything is fine and user can make map request
            Log.d(TAG, "isServicesOK: Google Play Services is working");
        }
        else if(GoogleApiAvailability.getInstance().isUserResolvableError(available)){
            //there is an error but we can fix it
            Log.d(TAG, "isServicesOK: An error is there but we can fix it");
            Dialog dialog = GoogleApiAvailability.getInstance().getErrorDialog(this, available, ERROR_DIALOG_REQUEST);
            dialog.show();
        }else{
            Toast.makeText(this, "We cant make maps request", Toast.LENGTH_SHORT).show();
        }
    }
}

在 Android 代码中，我刚刚添加了 if response.trim().equals("success")then 它将启动一个新活动。但是一直提示“登录失败”。

PHP脚本：

<?php
$conn =
 mysqli_connect("localhost","id7263189_root","dev@1234"
 ,"id7263189_logintb");
    mysqli_query($conn, "SET NAMES 'utf8'");

$username = $_POST["UserName"];
$password = $_POST["PassWord"];

$query = 'SELECT id FROM logintb WHERE username = "'.$username.'" and 
password = "'.$password.'"';
$result = mysqli_query($conn,$query) or die('error: ' .mysql_error());

if(mysqli_num_rows($result) == 1){

    echo "success";
}else{
    echo "error";
}


?>

目前我正在使用 XAMPP 连接到 myphpadmin。但是，如果我需要，我会将其托管到 000webhost 以使用真实设备与互联网进行检查。