title: Render link-field url and title separately in node template in Drupal 7
body: <p>If I print my link field in the node template like: [?php print render($content['field_link']); ?] I will get this field rendered the way it's set in the display settings of the content-type. </p>
  
  <p>But what if I want the url and the title of this field print out separately? I could print out a specific value in the array but that's not the way according to <a href="http://www.computerminds.co.uk/articles/rendering-drupal-7-fields-right-way" rel="nofollow">http://www.computerminds.co.uk/articles/rendering-drupal-7-fields-right-way</a></p>
  
  <p>Is there a simple way of doing this in Drupal 7. Thank you!</p>