PHP MYSQLI编写语句登录并检查用户状态

Question

我正在学习使用基于mysqli的视频教程制作网站.我开始知道使用预准备语句更安全,我正在尝试创建一个登录系统.这是我到目前为止所做的.

此代码可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();
 ?>

当我使用它时,我得到以下错误

警告:mysqli :: query()期望参数1为字符串,第33行的F:\ XAMPP\htdocs\login\login.php中给出的对象

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

我有数据库表列

user_id,用户名,密码(md5),user_level,status.

在user_level下我有以下内容

a = admin
m = member

在状态下

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

登录时我需要检查用户状态是否已激活它是否应该移动到索引页面,如果是,则应该显示用户已被停用,对其他用户也是如此.

如何在准备好的陈述中做到这一点？

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.');
}
?>

Answer 1

我想你需要看看mysqli_是如何工作的.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();