我的会话有问题，我的应用程序中有一个功能，该功能需要两种不同类型的用户登录，一个是管理员，这意味着他可以做所有事情在应用程序中可能是编辑器，其他可能只能在其中做几件事。
我的问题是会话似乎彼此重叠，我在管理员中登录，并且会话信息是来自编辑器的信息。
我从管理员登录了我：
<!---Login PHP--->
<?php
if( isset($_POST['btn-login']) ) {
$email = $_POST['email'];
$senha = $_POST['senha'];
$Error = false;
if (empty($email)) {
$Error= true;
$error = "Preencha o email.";
}
if (empty($senha)) {
$Error = true;
$error2 = "Preencha a senha.";
}
if($email) {
$sql = "SELECT email FROM admin WHERE email = '$email'";
$stmt = $conn->prepare($sql);
$stmt->execute();
$cout = $stmt->rowCount();
//echo "Email - ".$cout;
}
if($senha) {
$sql = "SELECT senha FROM admin WHERE email = '$email'";
$stmt = $conn->prepare($sql);
$stmt->execute();
$cout = $stmt->rowCount();
if($cout == 1) {
//echo "<br>Password - ".$cout;
$hashed = $stmt->fetch(PDO::FETCH_ASSOC);
//echo "<br>Password HASHED - ".$hashed['senha'];
$hashed_pass = $hashed['senha'];
}
}
if (!empty($email) && !empty($senha) && filter_var($email,FILTER_VALIDATE_EMAIL) && password_verify($senha,$hashed_pass) && !$Error) {
$sql = "SELECT email, senha FROM admin WHERE email ='$email' AND senha = '$hashed_pass'";
$query = $conn->prepare($sql);
$query->execute();
$count = $query->rowCount();
if($count == 1){
session_start();
$_SESSION['email'] = $email;
$_SESSION['senha'] = $crypt;
header("Location: home.php");
exit;
}
else {
$error = "Erro: password ou email errados";
}
}
}
?>
这是我的编辑登录：
<?php
/*EDITOR*/
if( isset($_POST['btn-login2']) ) {
$email = $_POST['email'];
$senha = $_POST['senha'];
if (empty($email)) {
echo "Preencha o email";
}
if (empty($senha)) {
echo "Preencha a senha";
}
if($email) {
$sql = "SELECT email FROM editor WHERE email = '$email'";
$stmt = $conn->prepare($sql);
$stmt->execute();
$cout = $stmt->rowCount();
//echo "Email - ".$cout;
}
if($senha) {
$sql = "SELECT senha FROM editor WHERE email = '$email'";
$stmt = $conn->prepare($sql);
$stmt->execute();
$cout = $stmt->rowCount();
if($cout == 1) {
//echo "<br>Password - ".$cout;
$hashed = $stmt->fetch(PDO::FETCH_ASSOC);
//echo "<br>Password HASHED - ".$hashed['senha'];
$hashed_pass = $hashed['senha'];
}
}
if (!empty($email) && !empty($senha) && filter_var($email,FILTER_VALIDATE_EMAIL) && password_verify($senha,$hashed_pass)) {
$sql = "SELECT email, senha FROM editor WHERE email ='$email' AND senha = '$hashed_pass'";
$query = $conn->prepare($sql);
$query->execute();
$count = $query->rowCount();
if($count == 1){
session_start();
// criar sessão com o email recebido por post e mandar o utilizador para a página home
$_SESSION['email_e'] = $email;
$_SESSION['senha_e'] = $senha;
header("Location: home.php");
exit;
}
else {
echo "Erro: password ou email errados";
}
}
}
?>
这是会话文件：
<?php
ob_start();
session_start();
// if session is not set this will redirect to login page
if( !isset($_SESSION['email']) && !isset($_SESSION['senha'])) {
header("Location: admin.php");
exit;
}
// ADMIN SESSIONS
if(isset($_SESSION['email'])){
//echo "entrei";
// select loggedin users detail
$res = "SELECT * FROM admin WHERE email='".$_SESSION['email']."'";
$stmt = $conn->prepare($res);
//echo "<br>SQL - > ".$res;
$stmt ->execute();
$count = $stmt ->rowCount();
if ( $count == 1 ) {
$userRow = $stmt->fetch(PDO::FETCH_ASSOC);
}
}
//EDITOR SESSIONS
if(isset($_SESSION['email_e'])) {
//echo "<br>Entrei2";
$sql = "SELECT * FROM editor WHERE email = '".$_SESSION['email_e']."'";
//echo "<br>SQL - > ".$sql;
$stmt = $conn->prepare($sql);
$stmt->execute();
$count = $stmt->rowCount();
if($count == 1) {
$userRow = $stmt->fetch(PDO::FETCH_ASSOC);
}
//echo "<br>Contagem - ".$count;
} else {
echo "<br>Sem Sucesso";
}
?>
当我试图解决这个问题时，我可能是因为我没有破坏会话，但仍然没有解决，我相信我可能在这里做错了。
注销文件：
<?php
session_start();
ob_start();
if (!isset($_SESSION['email']) || !isset($_SESSION['email_e'])) {
header("Location: index.php");
exit();
} else if(isset($_SESSION['email'])!="") {
header("Location: index.php");
exit();
}
//ADMIN LOGOUT
if (isset($_GET['logout'])) {
unset($_SESSION['email']);
unset($_SESSION['email_e']);
session_unset();
session_destroy();
header("Location: error.php");
exit;
}
ob_end_flush();
?>
事先感谢任何回答的人。
您正在通过创建2个不同的会话变量来处理两种用户，这似乎很奇怪。我认为正在发生的事情是，其中一个变量并没有得到不安，从而导致了您的问题。
使用相同的变量（$ _session ['email']），然后根据用户类型显示您想要的任何内容，这将变得更加简单和优雅。
认为您需要在某个时候添加一种新的用户类型：您必须编辑所有处理登录和注销的代码，这是不正常的。
尝试仅为管理员和编辑者创建一个登录页面，并根据其电子邮件从数据库中获取其用户类型。