自从我在做一个网络应用程序以来,我有一些问题,我对在不使用任何框架的情况下将 mvc 结构保留在网络应用程序中感到有点困惑。
这是我的应用程序的结构。
指数:
<!DOCTYPE html>
<head>
<meta charset="utf-8">
<!--<link rel="stylesheet" type="text/css" href="index.css" />-->
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.js"></script>
<script src="js/functions.js"></script>
<script>
$(document).ready(function() {
loginSend();
});
</script>
<title></title>
</head>
<body>
<div id="login">
Username: <input type="text" id="username" name="username"><br>
Password: <input type="password" id="password" name="password"><br>
<input id="send" type="button" value="Enviar">
</div>
</body>
</html>
loginsend 函数将数据发送到 service.php
<?php
include('server/main.php');
$action = $_POST['action'];
switch($action){
case "loginCheck":
$username=$_POST['username'];
$password=$_POST['password'];
$users = new users();
echo $users->loginCheck($username,$password);
break;
}
?>
该服务使用我在 main.php 中定义的一些类
<?php
class db{
public function conn() {
try{
$dbhost = "localhost";
$dbname = "eout";
$dbuser = "root";
$dbpass = "";
if($conn = new PDO("mysql:host=".$dbhost.";dbname=".$dbname, $dbuser, $dbpass)){
return($conn);
}
}
catch (Exception $e){
echo "Se ha presentado un error al conectar con la base de datos".$e;
}
}
}
class users{
function loginCheck($username, $password) {
try{
$db = new db();
$conn = $db->conn();
$pass_encriptada = md5 ($password);
$SQL_LOGIN_CHECK = "SELECT * FROM users where username='".$username."' and password='".$pass_encriptada."' and deleted='0'";
$conn->prepare($SQL_LOGIN_CHECK);
foreach($check = $conn->query($SQL_LOGIN_CHECK) as $row) {
$username_check = $row['username'];
if ($username_check == $username){
session_start();
$_SESSION['logged'] = 1;
$_SESSION['user_id'] = $row['id'];
$_SESSION['username'] = $row['username'];
$_SESSION['namelastname'] = $row['name'] ." ".$row['lastname'];
$usertype_id = $row['usertype_id'];
if ($usertype_id == 1 ){ //problema
$_SESSION['teacher'] = 1;
$response = "teacher";
$json = json_encode($response);
echo $json;
}
if ($usertype_id != 1 ){ //problema
$response = "user";
$json = json_encode($response);
echo $json;
}
}
}
}
catch(Exception $e){
echo "Se ha presentado un error en loginCheck".$e;
}
}
}
?>
在这种情况下,我理解 index.php 是我的观点,但是我对模型/控制器部分有一定的怀疑。我在我的示例中了解主文件是与数据交互的文件,所以它必须是模型......在我的情况下......它的 service.php 我的控制器?因为是与模型交互的那个?我对此感到有些困惑。万一我错了。如何制作控制器?
提前致谢。