How can I get a variable of a hash in php.
I have a variable on page like this
catalog.php#album=2song=1
How can i get the album and song values and put them into PHP variables?
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possible duplicate of Can PHP read the hash portion of the URL?Paul Alexander– Paul Alexander2011-12-07 18:00:28 +00:00Commented Dec 7, 2011 at 18:00
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3 Answers
You can't get this value with PHP because PHP processes things server-side, and the hash in the URL is client-side only and never gets sent to the server. JavaScript can get the hash though, using window.location.hash (and optionally call on a PHP script including this information, or add the data to the DOM).
Comments
Just adding onto @Alec's answer.
There is a parse_url() function:
Which can return fragment - after the hashmark #. However, in your case it will return all values after the hashmark:
Array
(
[path] => catalog.php
[fragment] => album=2song=1
)
As @NullUserException pointed out, unless you have the url beforehand this really is pointless. But, I feel its good to know nonetheless.
2 Comments
NullUserException
Unless you have the URL beforehand, this is useless.
Andelas
Thank you. Sounds like I could work this idea into what I need.
You can use AJAX/PHP for this. You can get the hash with javaScript and load some contents with PHP. Let's suppose we're loading the main content of a page, so our URL with hash is "http://www.example.com/#main":
JavaScript in our head:
function getContentByHashName(hash) { // "main"
// some very simplified AJAX (in this example with jQuery)
$.ajax({
url: '/ajax/get_content.php?content='+hash, // "main"
success: function(content){
$('div#container').html(content); // will put "Welcome to our Main Page" into the <div> with id="container"
}
});
}
var hash=parent.location.hash; // #main
hash=hash.substring(1,hash.length); // take out the #
getContentByHashName(hash);
The PHP could have something like:
<?php
// very unsafe and silly code
$content_hash_name = $_GET['content'];
if($content_hash_name == 'main'):
echo "Welcome to our Main Page";
endif;
?>
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