Catchable fatal error: Object of class PDO could not be converted to string on line 35

Question

我正在尝试以用户身份登录，但不知道收到此错误，但是当我第二次尝试时，我将可以访问，但登录后仍然如此，并且在第二次尝试登录后，我无法获取电子邮件从 DB 看来它没有创建 session.Its 只是传递代码而不创建会话。

if (!isset($_SESSION['email']) && isset($_POST['email'])) {
    if (!empty($_POST['email']) && !empty($_POST['password'])) {

        $email = filter_var($_POST['email'], FILTER_SANITIZE_STRING);
        $password = filter_var($_POST['password'], FILTER_SANITIZE_STRING);
        $password=sha1($password);

        try {
            $dbh = new PDO("mysql:host=$hostname; dbname=$database", $username, $pass);
            $dbh -> setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

            $email = $_POST['email'];
            $password = $_POST['password'];

            $stmt = $dbh -> prepare("SELECT email, password FROM tbl_user WHERE email = '".$email."' and password = '".$password."'");
            $stmt -> bindParam('".$email."', $email, PDO::PARAM_STR);
            $stmt -> bindParam('".$password."', $password, PDO::PARAM_STR);
            $stmt -> EXECUTE();
            $em = $stmt -> fetchColumn();

            if ($em == true) {
                // session_register("email");
                // session_register("password");

                $_SESSION['email'] = $_POST['email'];
                $_SESSION['START'] = time();

                setcookie("username", $_POST['email'], mktime()+(60*3), "/");
                setcookie("LUS", time(), mktime()+(60*3), "/");

                $stmt -> $dbh -> prepare("SELECT Name FROM tbl_user WHERE email = '".$email."'");
                $stmt -> EXECUTE();

                $em2 = $stmt -> fetchColumn();
                echo "Logged in.";
            } else {
                echo "email or password is incorrect.";
            }
        } catch (Exception $e) {
            echo "".$e->getMessage();
        }
    } elseif (empty($_POST['email']) && !empty($_POST['password'])) {
        # code...
        echo "Error : Enter your E-mail.";
    } elseif (!empty($_POST['email']) && empty($_POST['password'])) {
        # code...
        echo "Error: Enter your Password";
    } else {
        echo "Error: Enter your E-mail & Password";
    }

}
elseif (isset($_SESSION['email'])) {
    # code...
    echo "Welcome again you still logged in <strong>" .round((time() . $_SESSION['START'])/60) ."</strong> minutes(s) ago <a href='logout.php'>LogOut</a>";
}
elseif (!isset($_SESSION['email'])) {
    # code...
    echo "You must loggin first.";
    //header('location:../index.php');
}

Answer 1

更改第二个准备语句的用法

  $stmt -> $dbh -> prepare("SELECT Name FROM tbl_user WHERE email = '".$email."'");

至

  $stmt = $dbh -> prepare("SELECT Name FROM tbl_user WHERE email = '".$email."'");

Catchable fatal error: Object of class PDO could not be converted to string on line 35

Catchable fatal error: Object of class PDO could not be converted to string on line 35

php

mysql

pdo