screen.Width in PHP

Anon

Hi!

I'm trying to get the window size in PHP, so I want to make a layer (not in javascript) with the size minus 10 or 20, so the layer is "centered" on the screen.

I don´t know if I explain myself so clear :-/ (Sorry for my poor english). Anyway I looked in the manual and search in this phorum and found nothing about any "WindowSize" function in PHP. Any ideas?

Cheers!

Anon

I don't think you can! Usually javascript is the way to go when you want to extract userinformation.

Make a page with a javascript, get your info and go directly to your php-page or however you want it..

<?php
if(!$checked)
{
echo '<script language="Javascript">';

echo 'if(document.all) width = screen.Width;';
echo 'if(document.layers) width = "" /don't know..';
echo 'location.href = "page.php3?checked=true&width="+width;';
echo '</script>';
}

echo "Width: $width";

There may be some coding mistakes here, wrote it down quickly, this also not may be the best way..

Anon

I don't believe this is possible because PHP is run serverside and only has access to certain client information that is sent to the server. I don't believe windowsize is one such peice of information.

Your english is very understandable =)

harlan

Anon

Following code is currently a solution for you. Not that, if it does not run's in some browsers, default screen size is required!

<?

function get_screen_size() {
GLOBAL $screen_width;
GLOBAL $screen_height;
$screen_width = echo "<script>document.write(screen.Width);</script>";;
$screen_height= echo "<script>document.write(screen.Height);</script>";;
$screen_width = Chop(trim($screen_width));
$screen_height = Chop(trim($screen_height));
if (!$screen_width) $screen_width = 800;
if (!$screen_height) $screen_height = 600;
}
get_screen_size();
echo "screen_width=\"$screen_width\" AND screen_size=\"$screen_height\"";
?>