php - One file, displaying code on one page, but excluding on another? -

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can advise how i'd alter template called on every page load display particular section on 1 page?

i hope explain enough...

index.php includes;

require_once('./cache/templates/sidebar.php');

every subsequent page built uses what's defined in index.php file, meaning sidebar.php must.

i'm wanting edit sidebar.php contain advert displays solely on index page.

at moment, when edit sidebar.php, instance, display letter "b", display on homepage, , every other page so;

index page: http://img.photobucket.com/albums/v684/nilsatis/1stack.jpg every other page: http://img.photobucket.com/albums/v684/nilsatis/2stack.jpg

how can dictate 1 area of included file display on 1 page exclude showing on others?

edit: time. it's appreciated.

i inherited/purchased website , i'm finding index.php file temperamental.

i'm attempting put code within sidebar (or below it, above footer) amend on index.php breaks it.

        }          if (isset($url[1])) require_once('./cache/html/'.$url[0].'/'.$url[1].'.php');         else require_once('./cache/html/'.$url[0].'.php');     } } require_once('./cache/templates/sidebar.php');   }   require_once('./cache/templates/footer.php');

a quick , dirty hack inserting custom css stylesheet on index.php page makes div that's part of "sidebar.php" visible.

example:

sidebar.php:

  <?php     echo '<div id="theletterb" style="display:none">b</div>';   ?>

index.php:

  <?php     if ($_server['php_self'] == 'index.php') // replace differentiate between pages     {       echo '<style type="text/css"> #theletterb { display:block; } </style>';     }   ?>

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