How? Communicate login status FROM an included page

sherry

Hello.

Is there a way to communicate login details FROM an 'included' page back to the calling page?

For example:

Page 1 (a static page) calls Page 2's index.php page via an include() where Page 2 is a software that has a login feature on that page.

Once someone logs in, they are now navigating inside of Page 2's software and no longer on Page 1.

The issue is, if while inside of Page 2's software (any page) and someone clicks a link back to Page 1, Page 1 shows Page 2's login box as though that person is not logged in.

Once that person again clicks on any other link associated with Page 2's software, the index.php that was included which they are viewing from Page 1 since it is 'included', it again shows that person is still logged in.

So the question again is, how to get Page 1 to receive Page 2's login information so that when moving from Page 2 back to Page 1, the login box doesn't appear?

Whew...

There is a real reason why there is a link to Page 1 inside of Page 2's software but that is for another day. I'm just looking for the mechanics right now.

Thanks.

big_nerd

sherry,

You may want to look into using sessions in PHP.

A Session is a variable that can be passed around the site that contains information (in your case, login information).

As long as they are all PHP files and are being parsed as PHP (not HTML).

You can try various tutorials, like this one.

sherry

Thanks big.nerd.

I forgot about using sessions, but... would it be the username and password parameters only to define via sessions? I am using various software programs with the include() method not just one particular software.

Thanks.

sherry

Hello again.

I looked at the code for one particular software application. The username and password of Page 2 are NOT the 'final' parameters controlling whether the login box shows or not. It is the value of a particular session variable that is defined after the form is submitted.

Page 1, which has session_start(), is calling Page 2's page code via file_get_contents() not include(). That content is stored in a variable.

Prior to the creation of that variable, I double-checked, via print_r(), to make sure the session variable needed from Page 2 after that page was submitted, is present in Page 1's namespace and it is.  That's a good thing. 

I then use 'echo' on Page 1 to output Page 2's page code that is stored in that previously-mentioned variable.

The problem is, the login box still shows up on Page 1. It's like Page 2's code that is stored in that variable doesn't see the session var available on Page 1 prior to the 'echo' statement.  I used double-quotes in file_get_contents, also.

What might be the issue? 

Thanks.

Derokorian

try using eval($page2) instead of echo?

sherry

Thanks for your reply, Derokorian.

I tried eval($page2):

  $file_var1 = file_get_contents("http://www.webpage dot com");
  $file_var1 = eval($file_var1);
  echo "$file_var1";

but it gave the following error:

Parse error: syntax error, unexpected '<' in /foodir/foo/www/index.php(12) : eval()'d code on line 1

I then tried using Curl: e.g. curl_exec(...) et al.

The result still didn't have an effect on the login box show/hide var:

$show_login_box = ( ( $_SESSION['user_id'] == '' ) ? 1 : 0 );

Something else must be going on... 😕

========

I then found a way to see on Page 1, what the value of that hide/show login box was from Page 2 by creating a $_SESSION var.

To my surprise, that show/hide var IS being assigned properly, even when using file_get_contents(). :rolleyes:

PROBLEM PERSISTS: The login box is still there.

QUESTION: Could this be a COOKIE problem/solution? 😕

Via print_r($GLOBALS) for _COOKIE and HTTP_COOKIE, Page 2 has user, pass, and PHPSESSID. For Page 1, only PHPSESSID is present.

Thanks.