我可以将简单的表单发布转换为ajax表单发布吗

时间:2019-01-27 07:38:34

标签: php jquery mysql ajax

我想将此简单的投稿表格转换为ajax投稿表格

我尝试执行此操作,但无法执行此操作,请帮助我,下面给出了我的代码

我想将其更改为ajax提交,请您帮我

这是我的login.php

session_start();
require_once 'class.user.php';
$user_login = new USER();
if($user_login->is_logged_in()!="")
{
$user_login->redirect('../mydrive/dashboard');
  }
 if(isset($_POST['btn-login']))
{
  $email = trim($_POST['txtemail']);
  $upass = trim($_POST['txtupass']);
    if($user_login->login($email,$upass))
{
$user_login->redirect('../mydrive/dashboard');
}
  }
?>

class.user.php函数

public function login($email,$upass)
       {
        try
        {
$stmt = $this->conn->prepare("SELECT * FROM users 
WHERE email=:email_id");
            $stmt->execute(array(":email_id"=>$email));
            $userRow=$stmt->fetch(PDO::FETCH_ASSOC);

            if($stmt->rowCount() == 1)
            {
                if($userRow['userStatus']=="Y")
                {
                    $checkedupass = 
    $userRow['userPass'];

   if(password_verify($upass,$checkedupass))


                    {
            $_SESSION['userSession'] =  $userRow['id'];
                        return true;
                }
                else
                {

如果用户详细信息不正确或未登录或找不到帐户:

    header("Location: login? 
   error=invalid_login");
   exit;
                }
            }
            else
            {
                header("Location: login? 
  error=inactive");
                    exit;
                }   
            }
            else
            {
                header("Location: login? 
   error=AccountNotFound");
            exit;
            }       
        }
        catch(PDOException $ex)
        {
            echo $ex->getMessage();
        }
    }

0 个答案:

没有答案