Hi Jeff
 
  i saw ur code first of all see the correct code,  and try it.
 
  <?php
      Session_start();
      if ($_SESSION['auth'] != "yes")
      {
        echo  "<p><a href=''>Login /
  Signup</a></p>";
        echo  "<p><a href=''>Signout</a></p>";
          exit();
      }
  ?>
 
  above is the correct that should be written.
 
  here are the code errors in your code
 
  1. @ should not be used with $_SESSION variables.
  2. = !'yes' not correct, the correct code is !='yes'.
  3. page should be saved as .php extention like abc.php, i think u r 
using the php code in HTML page that's why it's not giving u any
output.
 
 
  try above option, i m sure it will solve ur probs.
  in case still u face any problem, i would live to see ur actual code
what is written on ur page.
 
  BABAL

--- In [email protected], "jnoelcook" <[EMAIL PROTECTED]> wrote:
>
> All,
>
> I am a newbie to php, and have many questions.
>
> I am trying to create a link to appear on an HTML page that
> shows: "Login / Signup" when not logged in, and: "Signout" when one
> IS logged in.
>
> I placed the following lines of code in my HTML text in the place
> where I want the link to be.
>
> <?php
>      
>       if (@$_SESSION['auth'] = !"yes")
>     {
>         echo = "<p><a href=''>Login /
> Signup</a></p>";
>         echo = "<p><a href=''>Signout</a></p>";
>             exit();
>       }
> ?>
>
> However, when I go to this page, no link whatsoever appears.  Could
> someone provide the correct code for this command?
>
> Thanks much!!!
>
> Jeff
>









The php_mysql group is dedicated to learn more about the PHP/MySQL web database possibilities through group learning.



SPONSORED LINKS
American general life and accident insurance company American general life insurance company American general life
American general mortgage American general life insurance Computer internet security


YAHOO! GROUPS LINKS




Reply via email to