Listing all slugs?

Question

I have a number of slugs and items in my custom category:

SLUG - number

video - 5 items

music - 5 items

[no-slug] - 5 items

empty - 0 items

blabla - 0 items

I want to list all non-empty slugs (otpionally with number of items next to them), like:

[ video (5) ] [ music (5) ] [ other (5) ]

(empty and blabla omitted)

Anyone knows how to do that?

All I know is how to get a slug for particular item:

<?php $slug = $terms[0]->slug; ?>

UPDATE

My taxonomy registration:

functions.php

register_taxonomy("op", array("portfolio"), array(
  "hierarchical" => true, 
  "label" => "Categories", 
  "singular_label" => "Category",  
  "rewrite" => true,
));

page-portfolio.php

$taxonomy = "op";
function get_taxonomy_list_html($taxonomy) {
  $term_links = array();
  foreach(get_terms('Category') as $term) {
    if (!empty($term->slug) && $term->count>0) {
      $link = get_term_link($term,$taxonomy);
      $term_links[] = "[ <a href='{$link}'>{$term->name}</a> ({$term->count}) ]";
    }
  }
  return implode(' ',$term_links);
}

Displays nothing :(

Answer 1

Hi @Wordpressor:

Sounds pretty simple so I'm concerned I misunderstand, but in case I did understand correctly here's a function; just pass the taxonomy to it like this: echo get_taxonomy_list_html('category'):

function get_taxonomy_list_html($taxonomy) {
  $term_links = array();
  foreach(get_terms($taxonomy) as $term) {
    if (!empty($term->slug) && $term->count>0) {
      $link = get_term_link($term,$taxonomy);
      $term_links[] = "[ <a href='{$link}'>{$term->name}</a> ({$term->count}) ]";
    }
  }
  return implode(' ',$term_links);
}