如何使用PHP从MySQL检索的用户名和密码打开另一个活动？

Question

使用PHP脚本在MySQL中确认用户名和密码后，如何在Android Studio中打开另一个活动？

我试图使用HTTP，但是由于我的SDK版本（28）而无法使用。所以，现在我正在尝试使用URL进行连接。

到目前为止，这是我的代码。

    package com.example.myapp;


    import android.content.Context;
    import android.os.AsyncTask; 
    import android.os.Bundle; 
    import android.support.v7.app.AppCompatActivity;
    import android.view.View;
    import android.widget.Button;
    import android.widget.EditText;
    import android.widget.TextView;
    import android.widget.Toast;
    import org.json.JSONArray;
    import org.json.JSONObject;
    import java.io.BufferedReader;
    import java.io.BufferedWriter;
    import java.io.InputStream;
    import java.io.InputStreamReader;
    import java.io.OutputStream;
    import java.io.OutputStreamWriter;
    import java.net.HttpURLConnection;
    import java.net.URL;
    import java.net.URLEncoder;
    import java.util.ArrayList;
    import java.util.HashMap;

    public class Sign extends AppCompatActivity {

    //Connector Variables

    private static final String jsonurl = "http://192.168.1.2/login2.php";
    private static TextView Info = null, Info2 = null, jSON;
    String username;
    String password;

    String usern[], userp[];
    int counter = 4;
    EditText pas;
    EditText usr;
    Button create_account;
    String ConnectionResult = "";
    Boolean isSuccess = false;
    private Button login;
    private Context text;
    private View view;
    private String LoginSMessage = "Login successful";
    String success = "succesful";


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

        Button login = findViewById( R.id.btnLogin );
        usr = ( EditText ) findViewById( R.id.etUsername );
        pas = ( EditText ) findViewById( R.id.etPassword );
        Info = ( TextView ) findViewById( R.id.tvInfo );
        jSON = ( TextView ) findViewById( R.id.tvJSON );

        Button create_account = findViewById( R.id.btnCreate );

        login.setOnClickListener( new View.OnClickListener() {

            @Override
            public void onClick(View v) {

                username = usr.getText().toString();
                password = pas.getText().toString();

                String type = "login2";
                GetJSON json = new GetJSON(this);
                json.execute(type, username, password);


            }
        });
    }


       class GetJSON extends AsyncTask<String, Void, String> {
       public GetJSON(Object o) {

       }

       protected String doInBackground(String... params) {

           String type = params[0];
           String login_url = "http://192.168.1.2/login2.php";
           BufferedReader bufferedReader = null;

           if (type.equals( "login2" )) {
               try {
                   String user_name = params[1];
                   String password = params[2];
                   URL url = new URL( login_url );
                   HttpURLConnection con = ( HttpURLConnection ) 
                   url.openConnection();

                   con.setRequestMethod( "POST" );
                   con.setDoOutput( true );
                   con.setDoInput( true );
                   OutputStream outputStream = con.getOutputStream();
                   BufferedWriter bufferedWriter = new BufferedWriter( new 
                   OutputStreamWriter( outputStream, "UTF-8" ) );
                   String post_data = URLEncoder.encode( "user_name", "UTF-8" 
                    ) + "=" + URLEncoder.encode( user_name, "UTF-8" ) + "&" + 
                   URLEncoder.encode( "password", "UTF-8" ) + "=" + 
                   URLEncoder.encode( password, "UTF-8" );
                   bufferedWriter.write( post_data );
                   bufferedWriter.flush();
                   bufferedWriter.close();
                   outputStream.close();
                   InputStream inputStream = con.getInputStream();
                   bufferedReader = new BufferedReader( new 
                   InputStreamReader( inputStream, "iso-8859-1" ) );

                   StringBuilder sb = new StringBuilder();
                   //bufferedReader = new BufferedReader( new 
                     InputStreamReader( con.getInputStream() ) );

                   String json;
                   while ((json = bufferedReader.readLine()) != null) {
                       sb.append( json + "\n" );
                   }
                   return sb.toString().trim();
               } catch (Exception e) {
                   return null;
               }
           }
           return null;


           }

            @Override
            protected void onPostExecute(String s){
           super.onPostExecute( s );
             jSON.setText(s);
                Log.d("CREATION", "Execution is done"+s);



               }


              }

我希望我的代码在确认MySQL中的用户名和密码后能够打开另一个活动。

Answer 1

您可以在服务器上编写以下PHP脚本：

$con=mysqli_connect("localhost","uname","pass","dbname");

   if (mysqli_connect_errno($con)) {
      echo "Error in connection : " . mysqli_connect_error();
   }

   $uname= $_POST['username'];
   $pass= $_POST['password'];
   $result = mysqli_query($con,"SELECT * FROM tableName where uname='$uname' and pass='$pass'");
   $row = mysqli_fetch_array($result);
   $data = $row[0];

   if($data){
      echo "Login success";
   }

   mysqli_close($con);

现在在您的android应用中，您可以有一个简单的if条件

@Override
protected void onPostExecute(String s){
  super.onPostExecute( s );
  jSON.setText(s);
  Log.d("CREATION", "Execution is done"+s);
  if(s!=null){
    Intent i = new Intent(your_current_activity_name.this,your_next_activity_name.class);
    startActivity(i);
  }else{
    Toast.makeText(getApplicationContext(), "Login credentials wrong", Toast.LENGTH_LONG).show();
  }
}