我无法弄清楚我做错了什么。我一直得到同样的错误

时间:2016-04-28 08:17:26

标签: php mysql

我一直收到这个错误。 '注意:尝试在第21行的C:\ xampp \ htdocs \ Homework \ Sessions \ Exercises \ Includes \ Login.php'中获取非对象的属性 这是我的代码如下。我正在尝试连接并登录到mysql数据库,但我一直遇到同样的错误。

有什么想法吗?感谢

<?php
    $dbEntries = $_POST;
    foreach ($dbEntries as &$entry) {
        $entry = dbString($entry);
    }

    @$db = new mysqli('mysql13.ezhostingserver.com', 'webdata', 'W3bD@t@', 'allicense');

    if (mysqli_connect_errno()) {
        echo 'Cannot connect to database: ' . mysqli_connect_error();
    } else {
        $query = "SELECT id AS EmployeeID, first_name AS FirstName, last_name AS LastName
                    FROM northwind_employees
                    WHERE email_address = '" . $dbEntries['Email'] .
                "' AND password = '" . $dbEntries['Password'] . "'";
        $result = $db->query($query);

        if ($result->num_rows) {
            $row = $result->fetch_assoc();
            $msg = 'Logged in as ' . $row['FirstName'] . ' ' . $row['LastName'];
            $_SESSION['FirstName'] = $row['FirstName'];
            $_SESSION['LastName'] = $row['LastName'];
            $_SESSION['EmployeeID'] = $row['EmployeeID'];
            if (isset($_POST['Remember'])) {
                setcookie('eid', $row['EmployeeID'], time() + 60 * 60 * 24 * 7);
            }
        } else {
            $msg = 'Login Failed';
            unset($_POST['LoggingIn']);
            $dbEntries = $_POST;
        }
    }
?>

0 个答案:

没有答案