Question

我的会话有问题，我的应用程序中有一项功能，需要两种不同类型的用户登录，一种是Admin，这意味着他可以在应用程序中尽一切可能，其他是编辑，他只能做一些事情。

我的问题是会话似乎互相重叠，我在管理员中登录，会话信息是编辑中的一个。

以下是管理员的登录信息：

<!---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";

            }

  }
  }

    ?>

继承我的编辑登录：

 <?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";

            }

  }
  }
  ?>

以下是Sessions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，我可能因为我没有破坏会话，但仍然没有解决这个问题，我可能在这里做错了我相信。

退出文件：

<?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();
?>

提前感谢任何人。

Answer 1

通过创建2个不同的会话变量，您正在处理这两类用户似乎很奇怪。我认为发生的事情是某种变量不会被取消，从而导致你的问题。

使用相同的变量（$ _SESSION [＆＃39; email＆＃39;]）然后根据用户类型显示您想要的任何内容会更加简单和优雅。

认为您希望在某个时候添加新的用户类型：您必须编辑处理登录和注销的所有代码，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，并根据他们的电子邮件从数据库中获取用户类型。

会话不起作用，相互重叠

1 个答案: