Question

这是因为我来自丹麦并使用谷歌翻译，因为我对英语不好，所以希望它能与之保持一致，但是，这就是我的MySQLI代码直到最后的话，以及说有错误。我曾尝试用正确的密码和电子邮件写入，但它不能以某种方式工作，它会一直犯错误，它怎么可能？

<?php
    session_start();

    include("include/database/db.php"); 

    if($stmt = $mysqli->prepare("SELECT id, djnavn, hemmelig, rank FROM `brugere` WHERE `email` = ? AND `password` = ?"))
            {
            $stmt->bind_param('ss', $email, $password);
            $email = $_POST['email'];
            $password = sha1($_POST['password']);
            $stmt->execute();
            $stmt->bind_result($id, $djnavn, $hemmelig, $rank);
            $stmt->fetch();
            $count_res = $stmt->num_rows;
            $stmt->close();

            if($count_res > 0) {
                $_SESSION["logged_in"] = true;
                $_SESSION["user_id"] = $id;
                $_SESSION["djnavn"] = $djnavn;
                $_SESSION["hemmelig"] = $hemmelig;
                $_SESSION["rank"] = $rank;
                if($rank == 0)
                {
                    echo "Your can not log in!";
                }
                if($rank == 1)
                {
                    echo "Ok, members you can log in now!";
                }
                if($rank == 2)
                {
                    echo "Ok, Admin you can log in now!";
                }
            }
            else {
                echo 'fail her: ' . $mysqli->error; 
            }
        }
?>

你可以帮帮我!!

Answer 1

首先是它的通知，因为您在使用之前没有定义$ email和$ password。

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

剪切并粘贴它们，因为你绑定参数：

$stmt->bind_param('ss', $email, $password);

说mysqli有错误

1 个答案: