Question

I have a button on a file test.php

<?php
$url="http://".$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'];
?>
<input type="button" name="btn" value="Download" onclick="alert('<?php echo $url; ?>')">

when the file is clicked, it alerts the page’s url.

I have another page, test2.php.

I want the url that is being alerted to be displayed out in test2.php

How can I accomplish this?

Answer 1

Depending on how important it is that it be a button, you could create a form that would send the $url in POST to test2.php. There are many workarounds, but one is the ever-elusive ‘hidden’ input.

The code on test.php would end up looking like:

<?php $url="http://".$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI']; ?>

<form action="test2.php" method="POST">
    <input type="hidden" name="url" value="<?php echo $url ?>" >
    <input type="submit" name="download" value="" onclick="alert('<?php echo $url; ?>')">
</form>

While the code on test2.php would have to include:

<?php $sent_url = $_POST['url']; 
      echo $sent_url;
?>

Hope this helps!

Mason

P.S. More information on other ways to do this can be found in this previous question.