PHP MYSQLI编写语句登录并检查用户状态

我正在学习使用一些基于 mysqli 的视频教程制作网站。我开始知道使用准备好的语句更安全，我正在尝试创建一个登录系统。这是我到目前为止所做的。

这段代码帮助我完全登录成功。

<form action ="" method="post">

User Name:<br/>
<input type='text' name='username' />
<br/><br/>
Password:<br/>
<input type='password' name='password' />
<br/><br/>
<input type='submit' name='submit' value='login'>
</form>
<?php

if(isset($_POST['submit'])){
    $username = $_POST['username'];
    $password = md5($_POST['password']);
    $stmt = $con->prepare("SELECT username, password FROM users WHERE username=? AND  password=? LIMIT 1");
    $stmt->bind_param('ss', $username, $password);
    $stmt->execute();
    $stmt->bind_result($username, $password);
    $stmt->store_result();
    if($stmt->num_rows == 1)  //To check if the row exists
        {
            while($stmt->fetch()) //fetching the contents of the row

              {$_SESSION['Logged'] = 1;
               $_SESSION['username'] = $username;
               echo 'Success!';
               exit();
               }
        }
        else {
            echo "INVALID USERNAME/PASSWORD Combination!";
        }
        $stmt->close();
    }
    else 
    {   

    }
    $con->close();
?>

但我还需要检查用户是否尚未激活或已被禁止或停用。所以我又做了一个代码。

这是我制作的代码

<?php
if(isset($_POST['submit'])){
$username = $_POST['username'];
$password = md5($_POST['password']);
$stmt = $con->prepare("SELECT username, password FROM users WHERE username=? AND    password=? LIMIT 1");
$stmt->bind_param('ss', $username, $password);
$stmt->execute();
$stmt->bind_result($username, $password);
$stmt->store_result();
if($stmt->num_rows == 1)  //To check if the row exists
  {
 $result=$con->query($stmt);
            $row=$result->fetch_array(MYSQLI_ASSOC);
            $user_id= $row['user_id'];
            $status = $row['status'];
            if($status=='d'){
                echo "YOUR account has been DEACTIVATED.";
            }else{
                $_SESSION['Logged'] = 1;
                $_SESSION['user_id'] = $user_id;
                $_SESSION['username'] = $username;
                echo 'Success!';
               exit();
            }
        }
        else {
            echo "INVALID USERNAME/PASSWORD Combination!";
        }
        $stmt->free_result();
        $stmt->close();
        
    }
    else 
    {   

    }
 $con->close();
 ?>

当我使用这个时，我收到以下错误

警告：mysqli_query() 期望参数 1 为字符串，即第 33 行 C:\XAMPP\htdocs\login\login.php 中给出的对象

致命错误：在第 34 行对 F:\XAMPP\htdocs\login\login.php 中的非对象调用成员函数 fetch_array()

我有数据库表列

用户身份， 用户名， 密码（md5）， 用户级别， 地位。

在 user_level 下我有以下内容

a = admin
m = member

状态下

a = activated
n = not activated
d = deactivated
b = banned

登录时，我需要检查用户状态是否以及它是否被激活，它应该移动到索引页面，或者如果它被激活，它应该显示用户已被停用，对于其他人也是如此。

如何在准备好的语句中做到这一点？

我在所有页面都有这个 connect.php

?php
//error_reporting(0);
'session_start';
$con = new mysqli('localhost', 'username', 'password', 'database');
if($con->connect_errno > 0){
die('Sorry, We\'re experiencing some connection problems.');
}
?>

我认为您需要了解一下 mysqli_ 的工作原理。这应该会让你朝着正确的方向前进。

if(isset($_POST['submit'])){
    $username = $_POST['username'];
    $password = md5($_POST['password']);
    $user_id = 0;
    $status = ""

    $stmt = $con->prepare("SELECT user_id, username, password, status FROM users WHERE username=? AND password=? LIMIT 1");
    $stmt->bind_param('ss', $username, $password);
    $stmt->execute();
    $stmt->bind_result($user_id, $username, $password, $status);
    $stmt->store_result();
    if($stmt->num_rows == 1)  //To check if the row exists
        {
            if($stmt->fetch()) //fetching the contents of the row
            {
               if ($status == 'd') {
                   echo "YOUR account has been DEACTIVATED.";
                   exit();
               } else {
                   $_SESSION['Logged'] = 1;
                   $_SESSION['user_id'] = $user_id;
                   $_SESSION['username'] = $username;
                   echo 'Success!';
                   exit();
               }
           }

    }
    else {
        echo "INVALID USERNAME/PASSWORD Combination!";
    }
    $stmt->close();
}
else 
{   

}
$con->close();