I want to include all the files in one directory in a switch statement. SO far I have this:

PHP Code:

<?php
switch($_SERVER['PHP_SELF'])
{
     case '/travel.php':
          echo '<div id="top">';
          break;
     case '/sports.php':
          echo '<div id="top1">';
          break;
     case '/politics.php':
          echo '<div id="top2">';
          break;
     case '/index.php':
          echo '<div id="top3">';
          break;
     case '/submit.php':
          echo '<div id="top3">';
          break;
     case '/links.php':
          echo '<div id="top3">';
          break;
     case '/column.php':
          echo '<div id="top3">';
          break;
     case '/funny.php':
          echo '<div id="top3">';
          break;
default:
  echo '<div>';
}
?>

I want another case statement so that instead of including just one page like I have right now, it would include all the files in a directory such as 'travel'.

I don't know if this is very clear or not?

Thanks

Just to clarify, when you say you want to include files do you mean include them as in

PHP Code:

include('file.php'); 


If so your code would look something like this:

PHP Code:

$dir = opendir('/');

while($file = readdir($dir))
{
     $ext = strrchr($file, '.')
     if($ext == '.php')
     {
          include_once($file);
     }
}

closedir($dir); 


Nope.

Instead of

PHP Code:

<?php
switch($_SERVER['PHP_SELF'])
{
     case '/travel.php':
          echo '<div id="top">';
          break;
?>

I want something like

PHP Code:

<?php
switch($_SERVER['PHP_SELF'])
{
     case '/travel/...':
          echo '<div id="top">';
          break;
}

Where ''/travel/...'' is a directory. How would I make '<div id="top">' appear in the source code of all files within the directory 'travel'.

For example I want a page like http://mypage.com/travel/mexico.php to have '<div id="top"> in it's source code

For something like this I think it would be better to use dynamic content (pulling content from a database). Instead of using static files (ie /travel/mexico.php) use a single file that fetches the data (travel.php?location="mexico"). Any changes in travel.php would then effect each location.

If however, you want to stick to static files, you'll need to include a common file in each static page.

PHP Code:

<?php
$dir = str_replace('/', '', __DIR__);
switch($dir)
{
     case 'travel'
          echo '<div id="top">';
          break;
}
?>

Yeah. I want to use a database, and will prboably set one up in the future, but I don't know a lot about them, plus I'm using another company's host and I'm not sure how to connect to their mysql.

Since you brought that up. Could you show me some links on how to set up a flat file database,for my pages?

Thanks

If you have to use flat files look into XML
http://xtech06.usefulinc.com/schedule/paper/19

Ok. Sorry for not posting this earlier, but about your previous code. I modified it a little to suit my needs. It now looks like this...
<?php
$dir = str_replace('/', '', __DIR__);
switch($dir)
{
case 'travel':
echo '<div id="top">';
break;
case 'sports':
echo '<div id="top1">';
break;
case 'politics':
echo '<div id="top2">';
break;
case 'submit':
echo '<div id="top3">';
break;
case 'links':
echo '<div id="top3">';
break;
case 'column':
echo '<div id="top3">';
break;
case 'funny':
echo '<div id="top3">';
break;
default:
echo '<div>';
}
?>

It doesn't work though, and I'm not sure why. No errors showed up, it just doesn't display the <div> tags. I also have some code before it

<?php
switch($_SERVER['PHP_SELF'])
{
case '/travel.php':
echo '<div id="top">';
break;
case '/sports.php':
echo '<div id="top1">';
break;
case '/politics.php':
echo '<div id="top2">';
break;
case '/index.php':
echo '<div id="top3">';
break;
case '/submit.php':
echo '<div id="top3">';
break;
case '/links.php':
echo '<div id="top3">';
break;
case '/column.php':
echo '<div id="top3">';
break;
case '/funny.php':
echo '<div id="top3">';
break;
default:
echo '<div>';
}
?>

I'm not sure if this has anything to do with its problem. Thanks

echo $dir and post the output

Do you want me to do something like this?

PHP Code:

<?php
$dir = str_replace('/', '', __DIR__);
echo '' . $dir;
switch($dir)
{
case 'travel':
echo '<div id="top">';
break;
case 'sports':
echo '<div id="top1">';
break;
case 'politics':
echo '<div id="top2">';
break;
case 'submit':
echo '<div id="top3">';
break;
case 'links':
echo '<div id="top3">';
break;
case 'column':
echo '<div id="top3">';
break;
case 'funny':
echo '<div id="top3">';
break;
default:
echo '<div>';
}
?>

I tried this but it didn't work

He meant for you to echo the $dir variable and post the result here on the forum, so we may take a look at it.

ok when I did this I got the result '__DIR__' at the top of the page

My mistake. __DIR__ only works in PHP 5.3.

Replace __DIR__ with dirname(__FILE__)

Ok. I figured out the problem. I have the logo, and menu all in and external file that I include in every page. So when I echo dirname(__FILE__) It comes up with
'/usr/local/httpd/www.konetchreport.com/external' which is where I store the external file. Is there any way to get around this?