java - PHP 和 AlertDialog 在 Android 上不同步

Question

我目前正在使用 YouTube 上的教程构建一个简单的登录 Android 应用，该应用从数据库中获取数据并让用户登录。但是，当我点击登录按钮时，无论输入什么，它都告诉我登录已成功。下面是输入数据的 Java 代码和连接数据库的 PHP 代码，每次单击登录时，都会弹出警报对话框，提示“登录成功”。如果用户输入的数据不正确，我希望它显示“登录未成功”，有人可以告诉我怎么做吗？谢谢

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 = "";
        if (type.equals("Login")) {
            try {
                String userName = 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("userName", "UTF-8")+"="+URLEncoder.encode(userName, "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);
    }
}

====

<?php
require "conn.php";

$userName = $_POST["userName"];
$password = $_POST["password"];

$mysql_qry = "select * from database where userName like '$userName' and password like '$password';";

$result = mysqli_query($conn, $mysql_qry);

if(mysqli_num_rows($result) > 0) {
    echo "Successful login";
}
else{
    echo "Login not successful";
}

Answer 1

在你的 conn.php 文件中

$dbname = "";
$username = "";
$password = "";
$server = "";

这不应该为空。

填写适当的数据