我收到了这个登录表单,然后由users.php文件进行验证,但即使用户名和密码也很难在数据库上,它说密码错误。数据库工作正常,并且已连接到页面。
登录表单:
<?php
require_once('/funcoes/Users.php');
if( !empty($_POST)){
try{
$users = new OOP_Users();
$result = $users->login($_POST['username'], $_POST['password']);
if ( $result == 'ok'){
}
else{
echo "Username/Password erradas!!";
}
}
catch(Exception $e){
echo "Authentication error!";
}
}
else{
session_start();
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Login</title>
</head>
<body><?php
if(!isset($_SESSION['username'])){
?>
<form name="Formulario_Login" action="#" method="POST"/>
<table>
<tr>
<td> Username </td><td><input name="username" type="text" <span style="font-size: 15px;" ></input></td></tr>
<tr>
<td> Password </td><td><input name="password" type="password" <span style="font-size: 15px;"></input></td></tr></tr>
<td colspan="7"><input type="submit" value="Login"></td></tr>
</table>
<?php
}
else{
echo 'Welcome ' . $_SESSION['username'] . "! <br>".'<a href="logout.php">Logout</a>';
echo '<br><a href="editarUser.php">Alterar perfil</a>';
echo '<br><a href="apagarUser.php">Apagar conta</a>';
}
?>
</body>
</html>
在users.php中我得到了这个功能
Users.php:
public function login($username, $password){
$result = $this->_myDB->performQuery("SELECT * FROM `users` WHERE `username' = '$username' AND 'password' = '$password'");
if ( $result->num_rows != 1 ){
return ('Authentication error');
}
else{
session_start();
$_SESSION['username'] = $username;
return('ok');
}
}
即使很难,我也有用户名&#34; ines&#34;和密码&#34; ines123&#34;在数据库上,每次我尝试登录时都会收到错误:&#34;身份验证错误!&#34;
谁能告诉我,我做错了什么?