java - 无法创建 Android PHP MySQL 登录应用程序

Question

我正在创建一个 Android PHP MySQL 登录应用程序，它将检查来自 MySQL 服务器的用户名和密码。我从 here 获得教程

我已经检查了 MySQL 中的名称和密码，它与用户类型相匹配，但我仍然无法进入 HomePage Activity ，因为它显示 Invalid用户名或密码。

 private void login(final String username, final String password) {

        class LoginAsync extends AsyncTask<String, Void, String> {

            private Dialog loadingDialog;

            @Override
            protected void onPreExecute() {
                super.onPreExecute();
                loadingDialog = ProgressDialog.show(MainActivity.this, "Please wait", "Loading...");
            }

            @Override
            protected String doInBackground(String... params) {
                InputStream is = null;
                List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
                nameValuePairs.add(new BasicNameValuePair("username", username));
                nameValuePairs.add(new BasicNameValuePair("password", password));
                String result = null;

                try{
                    HttpClient httpClient = new DefaultHttpClient();
                    HttpPost httpPost = new HttpPost(
                            "http://192.168.1.7:80/Android/CRUD/login.php");
                    httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

                    HttpResponse response = httpClient.execute(httpPost);

                    HttpEntity entity = response.getEntity();

                    is = entity.getContent();

                    BufferedReader reader = new BufferedReader(new InputStreamReader(is, "UTF-8"), 8);
                    StringBuilder sb = new StringBuilder();

                    String line = null;
                    while ((line = reader.readLine()) != null)
                    {
                        sb.append(line + "\n");
                    }
                    result = sb.toString();
                } catch (ClientProtocolException e) {
                    e.printStackTrace();
                } catch (UnsupportedEncodingException e) {
                    e.printStackTrace();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                return result;
            }

            @Override
            protected void onPostExecute(String result){
                String s = result.trim();
                loadingDialog.dismiss();
                if(s.equalsIgnoreCase("success")){
                    Intent intent = new Intent(MainActivity.this, HomePage.class);
                    //intent.putExtra(USER_NAME, username);
                    finish();
                    startActivity(intent);
                }else {
                    Toast.makeText(getApplicationContext(), "Invalid User Name or Password", Toast.LENGTH_LONG).show();
                }
            }
        }

        LoginAsync la = new LoginAsync();
        la.execute(username, password);

    }

登录.php

<? php
require_once("dbConnect.php");

$con = mysqli_connect(HOST,USER,PASS,DB);

$username = $_POST['name'];
$password = $_POST['password'];

$sql = "select * from users where name='$username' and password='$password'";

$res = mysqli_query($con,$sql);

$check = mysqli_fetch_array($res);

if(isset($check)){
echo 'success';
}else{
echo 'failure';
}

mysqli_close($con);
?>

Answer 1

$username = $_POST['name'];

改成这个:

$username = $_POST['username'];