PHP MYSQLI编写的语句登录并检查用户状态

我正在学习制作一些基于mysqli的视频教程网站。我知道使用预先准备好的语句更安全，我正在尝试创建一个登录系统。以下是我迄今为止所做的工作

此代码帮助我完全成功登录

<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();
?>

地位低下

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

在登录时，我需要检查用户状态，如果它被激活，它应该移动到索引页，或者如果它是d，它应该显示用户已被停用，同样对于其他人

如何在准备好的报表中执行

我在所有页面中都有这个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.');
}
?>

---

我想你需要看看mysqli是如何工作的。这会让你走上正确的方向

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();

---

错误消息中有什么不清楚的地方？什么是

$result=$con->query（$stmt）用于？我不知道该怎么办我是斯图克我想知道我走对了吗不，你走错了路。出于某种原因，您选择的是完全无用的登录名和密码，但不是您需要的激活内容。我在使用mysql的教程中看到了这一点，并认为它将帮助我检查用户状态。我做错了吗？？？为什么投反对票呢？？？非常感谢你，先生。你是世界上第一个帮助我的人，而当我请求帮助时，其他人只是关上了我的门。当我看到它工作时，我流下了眼泪。先生，你会长寿的。是的，它起作用了。你不知道它是否正确？？？无论如何，非常感谢你，先生。这就是我所要求的。我知道逻辑是正确的，我只是在这里写代码时会犯一些小错误。
?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.');
}
?>

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();