How do i make script, that detects its URL and displays it? [closed]

Question

I have a facebook like button on my website, but the data-href attribute needs a link to like:

<div class="fb-like share-button fbbutton" data-href="htp://www.example.com/article/1.html" data-send="false" data-layout="button_count" data-width="450" data-show-faces="true"></div>

I have a lot of pages, and would like to copy the code on all my pages, and the code would automatically detect the url it's in. Example: data-href="$link$" so that i don't need to change the link to that of my pages.

Question: How do i make script, that detects its URL and displays it?

note: Website is written in html5

Answer 1

According to the Facebook developers guide, the Like button defaults to the current URL. So all you should need to do is remove the data-href attribute and it the same code will always use whatever page the user is on.

On the above page, just leave the URL box blank then click the "Get Code" button and follow the instructions.

Answer 2

In PHP:

<php
//pull this using whatever method you like like a DB or URL-parser
//$url = 'http://'.$_SERVER[HTTP_HOST].$_SERVER[REQUEST_URI]; //or some derivative
$url = "http://www.example.com/page.html"
?>

<div class="fb-like share-button fbbutton" data-href="<?php echo $url;?>" data-send="false" data-layout="button_count" data-width="450" data-show-faces="true"></div>

Answer 3

Normally, I would suggest this be done server-side instead of by scripting, as suggested by @ionFish.

If you want to do this on the client-side, then you can simply use the Javascript Location object. The all you need is to invoke window.location to get the URL.

In the simplest form you just need:

<script>
document.write(window.location);
</script>

Where you want to output the URL. There are probably better ways to write this, depending on how you generate the surrounding code.