I'm in the process of writing a custom views filter handler. In order to learn from the existing filters, I'd like to know how I can tell which filter class views is using when I'm configuring an (exposed) filter on the GUI.

1 Answer 1


Adding an exposed view filter inserts the following (sample) code into the view output:

 <div class="view view-... view-id-... view-display-id-default view-...-id-1">
    <div class="view-header">

    <div class="view-filters">
        <form action="..." accept-charset="UTF-8" method="get" id="views-exposed-form-...">
            <div class="views-exposed-form">
               <div class="views-exposed-widgets clear-block">
                  <div class="views-exposed-widget views-widget-filter-tid">
                     <label for="edit-tid">...</label>
                     <div class="views-widget">
                        <div class="form-item" id="edit-tid-wrapper">
                           <select name="tid" class="form-select" id="edit-tid" >
                              <option value="All">...</option>...
                 <div class="views-exposed-widget views-submit-button">
                    <input type="submit" id="edit-submit-..." class="form-submit" />

It is not clear, though, why you can't find out the same by simply creating a view, exposing its filter and looking up the source code. Perhaps you could explain your question better?

  • 1
    I think BetaRide is trying to create a custom filter in a module and then trying to implement a class into the field's output. So the question really is, how to define the class that gets attached to the HTML output through the module. That's my guess though.
    – Ambidex
    Jun 5, 2012 at 11:26

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