Checking URL

[rockinaway]

I want to check the URL, and if it is the one I am looking for, then I want to show something. So how would I check the URL?

[Broniukas]

you can open a file handled in server, if it's exist - server online .

 

or use fsockopen()

 

$fp = fsockopen("www.example.com", 80, $errno, $errstr, 30);

 

[rockinaway]

No you see I have my homepage, but it will also show on other pages of my website, where I don't want it to..

 

So I want to check the website address the user is currently at, and if it is the homepage, then I want it to display the homepage, if not then I don't want to show anything.. is this possible with PHP?

[Broniukas]

 

$_SERVER['PHP_SELF']

$_SERVER['SERVER_NAME']

 

 

check this for more $_SERVER variables

 

http://lt2.php.net/reserved.variables

[rockinaway]

Right if I use $_SERVER['PHP_SELF'] on a page it shows index.php ..

 

But the problem is I have 2 locations...

 

- index.php

 

- index.php?cat=12 ...

 

I only want to show it for the index.php and not the cat=12 one.. can I limit this?

[chigley]

<?php

if(!isset($_GET['cat'])) {

// Code here for when the page is accessed without ?cat= on the URL

}

?>

[rockinaway]

Sorry didn't explain it correctly..

 

Any page without index.php only should not be able to show.. isn't there a global sign I can use to symbolise anything after the .php?

 

So index.php should so it.. but, for example, index.php?f=23, index.php?g=343, index.php?h=434 all should not..

[chigley]

<?php

foreach($_GET as $k => $v) {
$get = true;
}

if($_SERVER['PHP_SELF'] == "index.php" && !$get) {
// Display code
}

?>

 

There's probably a cleaner way but that will work

[Daniel0]

<?php
if(count($_GET) != 0)
{
die();
}

// rest of code...
?>

[rockinaway]

It has all gone now :S

 

 

EDIT - my bad .. IT WORKS WELL GOOD! .. Even sorted other problems I had ... thanks Daniel....