说有错误Mysqli - says there are errors mysqli

says there are errors mysqli

本文关键字：Mysqli 有错误 | 更新日期: 2023-09-27

这是我来自丹麦并使用谷歌翻译，因为我的英语不好，所以希望它可以与水平，但是，这就是我的MySQLi代码直接下降到最后一个单词的方式，并说有错误。我试图写正确的密码和电子邮件，但它根本懒得以某种方式工作，它不断犯错误，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; 
            }
        }
?>

你能帮我吗！！

首先，

它是通知，它的发生是因为您在使用它们之前没有定义$email和$password。

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

剪切并粘贴它们，以便您绑定参数：

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