Question

首先，我想说的是我对开发android非常新，现在我只是按照如何开发一个也使用登录的应用程序的教程。

我已经完成检查用户是否存在，应用程序将显示一个状态，说明登录是否成功。

现在我想检查用户是否具有管理员角色或保存在数据库中的用户角色。

tqvm in advanced。

下面是编码

1. MainActivity.java

package com.example.user.mysqldemo;

import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.view.Menu;
import android.view.MenuItem;
import android.view.View;
import android.widget.EditText;

public class MainActivity extends AppCompatActivity {

    EditText UsernameEt, PasswordEt;
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        UsernameEt = (EditText) findViewById(R.id.etUserName);
        PasswordEt = (EditText) findViewById(R.id.etPassword);
    }

    public void OnLogin(View view){
        String username = UsernameEt.getText().toString();
        String password = PasswordEt.getText().toString();
        String type = "login";


        BackgroundWorker backgroundWorker = new BackgroundWorker(this);
        backgroundWorker.execute(type, username, password);
    }

}

2.BackgroundWorker.java

public class BackgroundWorker extends AsyncTask<String,Void,String> {
Context context;
AlertDialog alertDialog;
BackgroundWorker (Context ctx) {
    context = ctx;
}
@Override
protected String doInBackground(String... params) {
    String type = params[0];
    String login_url = "http://ipaddress/foldername/login.php";
    if(type.equals("login")) {
        try {
            String user_name = params[1];
            String password = params[2];
            URL url = new URL(login_url);
            HttpURLConnection httpURLConnection = (HttpURLConnection)url.openConnection();
            httpURLConnection.setRequestMethod("POST");
            httpURLConnection.setDoOutput(true);
            httpURLConnection.setDoInput(true);
            OutputStream outputStream = httpURLConnection.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 = httpURLConnection.getInputStream();
            BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream,"iso-8859-1"));
            String result="";
            String line="";
            while((line = bufferedReader.readLine())!= null) {
                result += line;
            }
            bufferedReader.close();
            inputStream.close();
            httpURLConnection.disconnect();
            return result;
        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    return null;
}

@Override
protected void onPreExecute() {
    alertDialog = new AlertDialog.Builder(context).create();
    alertDialog.setTitle("Login Status");
}

@Override
protected void onPostExecute(String result) {
    alertDialog.setMessage(result);
    alertDialog.show();
}

@Override
protected void onProgressUpdate(Void... values) {

    super.onProgressUpdate(values);
}

}

Answer 1

您login.php脚本必须返回登录状态的值以及响应中此人的角色。像这样的JSON结构，

{
    "status" = 1,   //0 or 1 depending on login result
    "role" = 1      //a number based on the user type
}

然后，您可以在onPostExecute中解析此JSON结果并显示结果。

@Override
protected void onPostExecute(String result) {
    JSONObject jso = new JSONObject(result);
    int status = jso.getInt("status")
    int role = jso.getInt("role")
    if(status == 1){
        alertDialog.setMessage(result);
        alertDialog.show();

        if(role == 0){
            //User
        } else if(role == 1){
            //Admin
        }
    } else{
        //Login Failed
    }
}

android应用程序检查数据库

1 个答案: