Question

我正在尝试使用电子邮件登录＆amp;在下面的代码的帮助下，表单内的密码，但我得到：Fatal error: Using $this when not in object context in

表格的

<form method="post">
<?php
if(isset($_GET['error']))
{
?>
<div>
<button data-dismiss='alert'>&times;</button>
<strong>Wrong Details!</strong> 
</div>
<?php
}
?>
    <input type="email" name="txtemail" />
    <input type="password" name="txtpass" />
    <button type="submit" name="btn-login">Sign in</button>
    </form>

isset

if(isset($_POST['btn-login']))
{   
    try 
  {
    $stmt = $this->conn->prepare("SELECT * FROM tbl_users WHERE userEmail=:email_id");

    $stmt->execute(array(":email_id" => $email));
    $userRow  = $stmt->fetch(PDO::FETCH_ASSOC);
        $password = password_hash('upass', PASSWORD_DEFAULT);

    if ( $stmt->rowCount() == 1 )
    {
      if ( $userRow[ 'userStatus' ] == "Y" )
      {
             if( password_verify(  $_POST[ "upass" ] , $userRow[ 'userPass' ] ) )
        {
            if( password_needs_rehash( 'PASSWORD', PASSWORD_DEFAULT ) )
            {
                $new_pass = password_hash('upass', PASSWORD_DEFAULT);
                // Update database
            }
          $_SESSION[ 'userSession' ] = $userRow[ 'userID' ];
          return true;
        }
        else
        {
          header( "Location: index.php?error" );
          exit;
        }
      }
      else
      {
        header( "Location: index.php?inactive" );
        exit;
      }
    }
    else
    {
      header( "Location: index.php?error" );
      exit;
    }
  }
  catch ( PDOException $ex )
  {
    echo $ex->getMessage();
  }
}

所以我从行中删除了$this->：

$stmt = $this->conn->prepare("SELECT * FROM tbl_users WHERE userEmail=:email_id");

＆安培;替换为

$stmt = conn->prepare("SELECT * FROM tbl_users WHERE userEmail=:email_id");，

现在我得到Parse error: syntax error, unexpected '->' (T_OBJECT_OPERATOR)

所以我尝试了：$conn->prepare

现在我面临：Fatal error: Call to a member function prepare() on a non-object in

所以我似乎需要更换旧代码来修复它，如果是这样如何解决这个错误：致命错误：在不在对象上下文中使用$ this

使用数据库连接文件更新了问题：

<?php
global $db;
$db = isset($_POST['db']) ? $_POST['db'] : '';
class Database
{

    private $host = "localhost";
    private $db_name = "designer5";
    private $username = "root";
    private $password = "645644";
    public $conn;

    public function dbConnection()

    {   
    global $db;
    $db = isset($_POST['db']) ? $_POST['db'] : '';
        $this->conn = null;    
        try
        {

            $this->conn = new PDO("mysql:host=" . $this->host . ";dbname=" . $this->db_name, $this->username, $this->password);
            $this->conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);  
        }
        catch(PDOException $exception)
        {
            echo "Connection error: " . $exception->getMessage();
        }

        return $this->conn;
    }
}
?>

注意：我在互联网上研究了很多链接，但我在任何地方找到了解决方案： - （

致命错误：当不在pdo中的对象上下文中时使用$ this

0 个答案: