开发者

When I successfully login an extra function.session-start is displaying and how to check session for sign-in-action page

开发者 https://www.devze.com 2022-12-18 03:23 出处:网络
I have a sign-in-action-form page which accept username and password from sign-in-form. Now most of things are working fine like on sign-out if I press back button logout display and session is destro

I have a sign-in-action-form page which accept username and password from sign-in-form. Now most of things are working fine like on sign-out if I press back button logout display and session is destroyed. But I am facing a very strange problem. When I login and if login is successful a link function.session-start is shown on this page. I have not make any href like this I don't know from where this link come but it is disturbing my page. What is this error and how I can remove it. Second thing is there any method to check session for this page. I know there would be, but listen we are starting session and assigning value to session variable when our username and password are valid and if I apply checking for this page on the onload event it will show logged out because onload I have not started session done some thing like this. I don't know much about this. But I think I have explained my problem.

sign-in-action-form.php

                $con=mysql_connect("localhost","root","");
                if(!$con)
                {
                die('Could Not Connect:'.mysql_error());
                } 

                mysql_select_db("tcs",$con);

                $usr=$_POST["username"];                 //pick username from login page
                $pwd=hash('sha1',$_POST['password']);    //pick password from login page and use hash algorithm to encrypt it

                $query="select * from employee where Username='$usr' and Password='$pwd'";  //serch that single row in which both r found
                $result=mysql_query($query,$con);


                    if ($result) 
                    {

                                $row=mysql_fetch_array($result);

                        if (($row["Username"]==$usr) && ($row["Password"]==$pwd))
                        {
                                session_start(); 
                                $_SESSION['employee']['id']=$row['User Id'];
                                $_SESSION['employee']['username']=$row['Username'];


                                echo "<font color=red>"."<h3 align=center>"."Welcome ".$_SESSION['employee']['username']."</h3>"."</font>";
                                echo "<br />"."<a href='upload_file.php'>"."<font color='white'>"."<h4>"."Up-Load Files"."</h4开发者_StackOverflow>"."<font>"."</a>";
                                echo "<br />"."<br />"."<a href='list_files.php'>"."<font color='white'>"."<h4>"."List All Up-Loaded Files"."</h4>"."<font>"."</a>";

                        }

                        else
                        {
                                echo "Login Not Successfull";
                        }

                    }
}

else
{
echo 'Error! Username & Password were not sent!';
}

?>

</font>
<a  href="logout_file.php"><font color="white"><h3 align="right">Sign Out</h3></font></a>
<font color="white">

</body>
</html>




 session.php

    <?php
    session_start();
    if(!isset($_SESSION['employee']))
    {
        echo "Your are Logged Out";
        exit;
    }   

else
{
echo "<blink>"."Welcome Mr.".$_SESSION['employee']['username']."</blink>";
}
?>


Implement a middle step like, "You have been logged out, redirecting you to front-page"

0

精彩评论

暂无评论...
验证码 换一张
取 消

关注公众号