How to correctly write this code?

I am looking for the correct why to right this code. The code I have shows up with my website address in front and I just need this to be a _blank target link to open!

Shows up like this: http://www.mysite.com/folder/folder/page/http://www.linkineed.com
Just need: http://www.linkineed.com

Thanks,
Adam

<?php if($link != '') : ?>
	        <li class="link clear">
            <button class="button"><a target="_blank" href="<?php echo $link; ?>"><?php _e('Download File', 'author'); ?></a></button>
	        </li>
	        <?php endif; ?>

Code (markup):

 

please tell us the contents of $link and <?php _e('Download File', 'author'); ?>

 

Well, I would guess that the contents of the $link would be the website page directory and than the "<?php _e('Download File', 'author'); ?>" would be the text field in the post editor that I write, but I am not 100% sure about that.

 

where are u got $link variable ??

 

Well, I found this code on the same page, does this help?

<?php
      $caption = get_post_meta(get_the_ID(), 'portfolio_caption', TRUE);
      $link = get_post_meta(get_the_ID(), 'portfolio_link', TRUE);
?>

Code (markup):

 

yes, but not if we don't know the content (use print_r($caption) or print_r($link) to display the results)

but i think if you remove the $link part you'll get the correct writen domain name!

 

Change the code to

<?php if($link != '') : ?>
             <li class="link clear">
             <button class="button"><a target="_blank" href="<?php echo $link; ?>"><?php _e('Download File', 'author'); ?></a></button>
             </li>
             <?php endif; 
echo '<br />'.$link;
exit;
?>

PHP:

Then you can see what the link is. If it's not http://www.linkineed.com follow get_post_meta(get_the_ID(), 'portfolio_link', TRUE); back to see where it's being set to what you don't want.

 

You'll probably need basename(), or some fancy Regex. It all depends on what you have.

Also, I believe Rukbat's code could be changed from if($link != '') to if($link) and echo '<br />'.$link;exit; to exit('<br />'.$link);