Using url as a variable

Question

I normally use a GET variable i n the URL to determined what data to show in my template.

Today I am not passing a variable. I am including a file named inludes/leftmenu.php on every page but depending on the page name I want to show different data.

leftmenu.php looks like so:

$page = $_SERVER["REQUEST_URI"];
$page_name = get_page_name($_SERVER["REQUEST_URI"]);

   if ($page_name == "classes.php")
        {
<h2 class="left_h2">Class Schedules &amp; Info</h2>
<ul class="left_ul">
    <li>Immersion Programs</li>
    <li>Afternoon Programs</li>
    <li>Immersion Schedule</li>
    <li>Afternoon Schedule</li>
</ul>

    }else if($page_name == "about.php")
    {

<h2 class="left_h2">About Us</h2>
<ul class="left_ul">
    <li>About Us</li>
    <li>Photo Scrapbook</li>
</ul>
   }else if ($page_name == "news.php")
    {

<h2 class="left_h2">News &amp; Events</h2>
<ul class="left_ul">
    <li>News</li>
    <li>Events</li>
</ul>

  }

Sorry,I had two issues in my head while writing this post and I got them mixed up. But you actually both questions got answered. Thanks! — Denoteone
– Denoteone, Commented Feb 27, 2011 at 23:03

Alnitak · Accepted Answer · 2011-02-27 22:53:08Z

2

You could use a PHP array:

$pages = array(
  '/events.php' => 'includes/events_menu.php',
  '/news.php'   => 'includes/news_menu.php'
);

// lookup the appropriate include file
$uri = $_SERVER['REQUEST_URI'];
$include = $pages[$uri];

// produce a default page if the URI wasn't recognised
if (!isset($include)) {
  $include = 'includes/default.php';
}

include($include);

edited Feb 27, 2011 at 22:53

answered Feb 27, 2011 at 22:47

Alnitak

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2 Comments

Denoteone Over a year ago

Can $pages have multiple keys for each value. i.e. $pages = array('/events.php, /home_events.php, away_events.php ' => 'includes/events_menu.php');

Alnitak Over a year ago

no, you'd have to have duplicate keys all mapping individually to the same value.

Floern · Accepted Answer · 2011-02-27 22:47:24Z

1

I don't know if it's faster, but there are many ways to do this.

E.g. with an array:

$page_name = get_page_name($_SERVER['REQUEST_URI']);

$includes = array(
    'events.php' => 'includes/events_menu.php',
    'news.php' => 'includes/news_menu.php'
);

if(isset($includes[$page_name])) include($includes[$page_name]);
else die('this page does not exist');

answered Feb 27, 2011 at 22:47

Floern

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Comments

Mikhail · Accepted Answer · 2011-02-27 22:44:57Z

0

PHP polulates $_GET with variables passed in the URL. Though I do not recommend having a user-passed variable as a function of what your code includes.

an array is a better method, ofcourse, but you may also look into realpath() and make sure it's in DOCUMENT_ROOT /includes

answered Feb 27, 2011 at 22:44

Mikhail

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1 Comment

Denoteone Over a year ago

Sorry I asked my question incorrectly. Please see my edits above.

mailo · Accepted Answer · 2011-02-27 22:48:05Z

0

You can create array with paths and then use in_array function, which will reduce if statements.

answered Feb 27, 2011 at 22:48

mailo

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Comments

isotrope · Accepted Answer · 2011-02-27 22:48:56Z

0

If there's a menu for every single page, you could go with

include('includes/'.basename($page_name, ".php").'_menu.php');

answered Feb 27, 2011 at 22:48

isotrope

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2 Comments

RobertPitt Over a year ago

Ouch, I will not be taking this advise !

isotrope Over a year ago

Robert, if ever you have a moment, could you please let me know why this is such bad advice (for personal knowledge and better understanding)? Thanks. I guess the solution doesn't really apply at all to the edited question, anyway.

mario · Accepted Answer · 2011-02-27 22:50:02Z

0

You can use REQUEST_URI, but you have to keep in mind that it always starts with a / slash. Unless you apply basename() you therefore would have to write:

if ($page_name == "/events.php")

The if statements are a workable approach if you only need a few page names. Otherwise use an switch or an url->filename map (as an array).

If you are using a RewriteRule to map all incoming URLs to a single PHP script, you could also think about using PATH_INFO instead of REQUEST_URI.

answered Feb 27, 2011 at 22:50

mario

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