I'm using views to output a link. The user gets two fields in which to add the href and the text on the link. I wan't them to be able to put a class on the link as well. How can I do this?

In views I have set a fixed class to all those links, but I wan't to be able to add another class if I wan't to, from the backend.

  • you want to add class to anchor tag <a> ? Commented Feb 13, 2013 at 15:40
  • Yes that's what I want to do.
    – Johan Dahl
    Commented Feb 13, 2013 at 15:49
  • Are you only asking how to add the class to the Views output, or do you also want help adding the form element and storing the data for the classname?
    – Andy
    Commented Feb 13, 2013 at 21:34
  • Both. I'm not sure I understand, but I wan't a form element in the backend where the user can put some text that will become the class of the link.
    – Johan Dahl
    Commented Feb 14, 2013 at 8:07
  • OK. And currently you have a link field attached to an entity, and that's what the user edits, yeah? (Also, are you using Views to display a list of links attached to different entities, or are you using it as a convenient way to display the link attached to just one entity?)
    – Andy
    Commented Feb 14, 2013 at 11:17

1 Answer 1


You can do this on the server side with a Views field tpl.php file, or on the client side with jQuery.

For the server side method, go to your view edit -> Advaced -> Theme:Information -> chose your field you need to create a one template file (in your theme/templates folder):

For example: if your field name is title you will see something similar to this:

Field Content: Title (ID: title): - click on it and cope the code

views-view-field--YOURVIEWNAME--page--title.tpl.php -paste it here

Now you have 3 variables to play with

 * - $field: The field handler object that can process the input
 * - $row: The raw SQL result that can be used
 * - $output: The processed output that will normally be used.

Easy way - do it with str_replace function :D

<?php $output = str_replace('<a', '<a class="someclass"', $output); ?>
<?php  print $output; ?>

Also you can play around with($row) and create your own output. To easy development install devel module so you can dpm($row) and see what it contains. or you can print_r($row). have fun <3

Fastest way to do this with jQuery you can simple add class to your fields anchor tag.

$(document).ready(function() {
    $('.your-views-class-name .your-view-field a').addClass('someclass');

open firebug(int firefox) and find your view class name and field class name which contains anchor tag and add class to it. You can do this in YOURTHEME - scripts.js or any other js file you want.

  • 1
    Yeah I know I can do it with jQuery. It's just that if feels wrong doing it that way. I don't mind it for small fixes, but to use jQuery to add classes for lot's of content just feels wrong.
    – Johan Dahl
    Commented Feb 13, 2013 at 16:23
  • yea, i can agree with you, probabaly you will need to create template for that field and manage classes there, need some research about this Commented Feb 13, 2013 at 16:29
  • Yeah I heard you can create templates for views. But I don't understand how that will give me a backend field to input the class name. However, I'm new to Drupal and views in particular so I have no clue really.
    – Johan Dahl
    Commented Feb 13, 2013 at 16:33
  • updated my answer Commented Feb 13, 2013 at 16:55
  • I'm trying to implement this solution. Is it necessary to implement the jquery and template file or is only creating the template file enough? I'm asking this as I did the template file but not getting the good result...
    – Justme
    Commented Apr 19, 2016 at 6:44

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