html linking help

cyronuts

I have a mysql table, which holds two cols: page_name, which designates an html page name, and page_code, which designates the code associated with that page name.

I am looking on the forum right now, but can't seem to find, how to present the page when a link is clicked.

ie., if I have a link called 'splash', what is the code for when a user clicks on the link to get them to view the 'splash' html code?

I know this is probably very simple, but I can't seem to find a snippet. THANKS!

sarahk

check out thread 10207362

In your case you would query the database to get a page matching the code 'splash' and then call the page in one of the ways described.

cyronuts

Is this looking close...? It is not working though.

<html>
<head>
<title>Untitled Document</title>
</head>
<body bgcolor="#FFFFFF" text="#000000">

require_once("sc_fns.php");

// links to the different info to display
echo"<p><a href=\"?page=splash\">splash</a> <a href=\"?page=page_one\">Page One</a> <a href=\"?page=page_two\">Page Two</a></p>";

      // list of possible pages comes from database
      $cat_array=get_html_pages();

if(!empty($GET)) { extract($GET); }
else { extract($HTTP_GET_VARS); }

if ($HTTP_GET_VARS['page']=="html_page_name")
{
echo "yes, yes, yes";
}

?>
<p>below the text data</p>
</body>
</html>

Here is what the get_html_pages() function looks like:

function get_html_pages()
{
// query database for a list of categories
$conn = db_connect();
$query = "select html_page_name from html_pages";
$result = @($query);
if (!$result)
return false;
$num_cats = @mysql_num_rows($result);
if ($num_cats ==0)
return false;

$result = db_result_to_array($result);
return $result;
}