How to assign js var to php var

Question

I'm having a struggle with the php code below

<?php
$list = 0;
$screenWidth = '<script 
type="text/javascript">document.write(screen.availWidth);</script>';

if($screenWidth > 700){
   $list = 1;
}

echo $screenWidth; 
?>

The echo is printing the value of $screenWidth (screen width) correctly.

So it means the js works and is assigning the value from screen.availWidth to $screenWidth correctly but somehow $list is always 0 ! Even when the screen width is 1920

Thanks for your help!

Ceco

You reallize that php runs on server and JS in user browser window? — dfsq, May 13 '17 at 17:29
JS is client side, PHP is only on the server. https://softwareengineering.stackexchange.com/questions/171203/what-are-the-differences-between-server-side-and-client-side-programming — chris85, May 13 '17 at 17:29
Your variable is always the literal ``, it is never the integer that JS returns. — chris85, May 13 '17 at 17:31
PHP is used on the server to generate HTML. When it is done, server sends it to your broweser and **the php process gets killed**. And only then Javascript code runs on your web-browser. — tereško, May 13 '17 at 17:31
Maybe this thread is useful http://stackoverflow.com/questions/1504459/getting-the-screen-resolution-using-php — chris85, May 13 '17 at 17:32
No real/best way to do it. Only hack would be for javascript to store data in cookie and php reading data from cookie but, not the best. http://stackoverflow.com/questions/21620133/how-to-assign-javascript-variable-value-to-php-variable — Robert Rocha, May 13 '17 at 17:36
Anyway I didn't understand why echo is printing the correct value and why I cannot convert it to integer — Tzvetan Jordanov, May 13 '17 at 18:04

score 0 · Answer 1 · answered May 13 '17 at 18:10

0

Your code doesnt echo the Screen width, it echoes the js:

echo '<script  type="text/javascript">document.write(screen.availWidth);</script>';

And on client side, this JS runs and prints out the screen width ( document.write).

However, as $screenWidth is a String, comparing it to 700 is not really working...

answered May 13 '17 at 18:10

Jonas Wilms

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Ok I understand that my code prints the js output of screen.availWidth and it is a string but normally we can (int)$string ... I don't know ... Guys, do you understand what I am asking ? Why echo prints the value of $screenWidth which is the js output screen.availWidth but cannot convert it to integer and compare it to let say 700 – Tzvetan Jordanov May 13 '17 at 18:20

How to assign js var to php var

1 Answers1