WordPress获取术语，但删除某些过滤器（如果存在）

Question

I have an image gallery, which has some filter controls. Now they work fine, all is well. The only thing that I need to do is remove the featured filter from showing on the gallery page, as this is only used to help pull out images on the homepage.

Currently on my projects page, I have this creating the list items which are links to the filter slugs.

Is there anyway I can say, if featured don't display?

<?php
// Get the taxonomy
$terms = get_terms('filter');

// set a count to the amount of categories in our taxonomy
$count = count($terms);

// set a count value to 0
$i = 0;

// test if the count has any categories
if ($count > 0) {

    // break each of the categories into individual elements
    foreach ($terms as $term) {

        // increase the count by 1
        $i++;


        $feat = term_exists('featured', 'filter', 'project');
        if ($feat !== 0 && $feat !== null) {
            $feat .= "";
        }



        // rewrite the output for each category
        $term_list .= '<li><a href="javascript:void(0)" class="' . $term->slug . '">' . $term->name . '</a></li>';

        // if count is equal to i then output blank
        if ($count != $i) {
            $term_list .= '';
        } else {
            $term_list .= '';
        }
    }

    // print out each of the categories in our new format
    echo $term_list;
}
?>
               </ul>

Answer 1

I figured out something that managed to work for me:

This:

$terms = get_terms('filter');

To this:

$terms = get_terms( 'filter', array(
    'exclude'    => '6'
) );

The 6 being the ID of the slug within my taxonomy.

Simple excludes one of the categories appearing in the list, meaning no link for someone to click. Job done. Hope this might help someone else.