我正在尝试让android应用程序与mysql数据库通信。即使在清单中设置了正确的权限，我也会出现以下错误：
java.net.socketexception:套接字失败：android studio中的eacces（权限被拒绝）
以下是androidmanifest.xml

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
    package="com.example.biddingapp">

    <uses-configuration android:name="android.permission.INTERNET"/>
    <uses-permission android:name="android.permission.ACCESS_WIFI_STATE" />
    <uses-permission android:name="android.permission.CHANGE_WIFI_STATE" />
    <uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />
    <uses-permission android:name="android.permission.CHANGE_NETWORK_STATE" />

    <application
        android:allowBackup="true"
        android:icon="@mipmap/ic_launcher"
        android:label="@string/app_name"
        android:roundIcon="@mipmap/ic_launcher_round"
        android:supportsRtl="true"
        android:theme="@style/AppTheme">
        <activity android:name=".SecondActivity"></activity>
        <activity android:name=".MainActivity">
            <intent-filter>
                <action android:name="android.intent.action.MAIN" />

                <category android:name="android.intent.category.LAUNCHER" />
            </intent-filter>
        </activity>
    </application>
</manifest>

这里是mainactivity.java

public class MainActivity extends AppCompatActivity {

    private EditText username;
    private EditText password;
    private Button login;

    Connection con;

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

        username = (EditText)findViewById(R.id.username);
        password = (EditText)findViewById(R.id.password);
        login = (Button)findViewById(R.id.login);

        login.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                new MainActivity.checkLogin().execute("");
            }
        });
    }

    private class checkLogin extends AsyncTask<String, String, String>{

        String str = null;
        Boolean isSuccess = false;

        @Override
        protected void onPreExecute() {
            super.onPreExecute();
        }

        @Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);
        }

        @Override
        protected String doInBackground(String... strings) {
            con = connectionClass(ConnectionClass.user.toString(),ConnectionClass.password.toString(),ConnectionClass.database.toString(),ConnectionClass.server.toString());
            if(con == null){
                runOnUiThread(new Runnable() {
                    @Override
                    public void run() {
                        Toast.makeText(MainActivity.this,"Check Internet Connection",Toast.LENGTH_LONG).show();
                    }
                });
                str = "On Internet Connection";
            }
            else {
                try {
                    String sql = "SELECT * FROM users WHERE username = '" + username.getText() + "' AND password = '" + password.getText() + "' ";
                    Statement stmt = con.createStatement();
                    ResultSet rs = stmt.executeQuery(sql);

                    if (rs.next()) {
                        runOnUiThread(new Runnable() {
                            @Override
                            public void run() {
                                Toast.makeText(MainActivity.this, "Login Success", Toast.LENGTH_LONG).show();
                            }
                        });
                        str = "Success";

                        Intent intent = new Intent(MainActivity.this, SecondActivity.class);
                        startActivity(intent);
                        finish();
                    } else {
                        runOnUiThread(new Runnable() {
                            @Override
                            public void run() {
                                Toast.makeText(MainActivity.this, "Check email or password", Toast.LENGTH_LONG).show();
                            }
                        });

                        username.setText("");
                        password.setText("");
                    }
                } catch (Exception e) {
                    isSuccess = false;
                    Log.e("SQL Error : ", e.getMessage());
                }
            }
            return str;
        }
    }

    @SuppressLint("NewApi")
    public Connection connectionClass(String user, String password, String database, String server){
        StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build();
        StrictMode.setThreadPolicy(policy);
        Connection connection = null;
        String connectionURL = null;
        try{
            Class.forName("com.mysql.jdbc.Driver");
            connectionURL = "jdbc:mysql://localhost/biddingapp://" + server+"/" + database + ";user=" + user + ";password=" + password + ";";
            connection = DriverManager.getConnection(connectionURL);
        }catch (Exception e){
            Log.e("SQL Connection Error : ", e.getMessage());
        }

        return connection;
    }
}

下面是connectionclass.java

package com.example.biddingapp;

import android.annotation.SuppressLint;
import android.os.StrictMode;
import android.util.Log;

import java.sql.Connection;
import java.sql.DriverManager;

public class ConnectionClass {
    public static String user = "androidstudio";
    public static String password = "program";
    public static String database = "users";
    public static String server = "localhost";
}

如何解决这个问题？有没有更好的办法？