Using PHP Snippet to Display Taxonomy Terms as PAGE of Links (not within nav menu)

Create a page (or a story), place

<?php
$vid = 1;  // Set the vid to the vocabulary id of the vocabulary you wish to list the terms from
$items = array();

$terms = taxonomy_get_tree($vid);
foreach ( $terms as $term ) {
  $items[] = l($term->name, "taxonomy/term/$term->tid");
}

if ( count($items) ) {
  print theme('item_list', $items);
}
?>

in the body, make sure you set the $vid to the vocabulary you want to see.

Also. make sure the input format is 'PHP code' and you probably do not want the page promoted to the front page.

Have you tried the taxonomy_list module? I used it on a site of mine, and it does exactly what you need.

Nevets, thank you for your code.

I tried to modify the php to list hierarchical terms that have at least one node within them, but I cannot get it to work.

<?php
$vid = 5;  // Set the vid to the vocabulary id of the vocabulary you wish to list the terms from
$indent = "--";  // Set to what you want the indent to be
$items = array();

$terms = taxonomy_get_tree($vid);
foreach ( $terms as $term ) {
  $count = db_result(db_query("SELECT COUNT(nid) FROM {term_node} WHERE tid = %d", $term->tid));
  if($count) {    // kick out the empty terms
  $pole[]=Array (l($term->name, "taxonomy/term/$term->tid") . " ($count)", $term->depth, $count, $term->tid)  ;
    }
  print '<div>' . str_repeat($indent, $term->depth) . l($term->name, "taxonomy/term/$term->tid") . " ($count)"  . '</div>';
}
?>

Would anyone, please, attempt to correct this (or link to a place where the answer is already provided)? I looked and I looked, but I cannot find the solution.

This is a great thread. Thanks for the info and customizations. I'm looking for another variation if anyone has any ideas. I need to create a link around a box that is to the most current item posted in a single vocabulary id or of a certain content type.

So what I'm looking for is something that will let me do:

<a href="GENERATED-LINK-GOES-HERE">
Content I want to link is here but not the title from the node
</a>

Does that make sense?

Oh, and I'm on 6.x

Help MUCH appreciated!

Hi,

Am using the snippet, to show the count of number of nodes tagged with a term, how can I also show the count for total number of comments in the nodes.

The snippet I am using is:

unset ($output);

$vocabulary_id = 12;
$result = db_query("SELECT d.tid, d.name, MAX(n.created) AS updated, COUNT(*) AS count FROM {term_data} d INNER JOIN {term_node} USING (tid) INNER JOIN {node} n USING (nid) WHERE d.vid = $vocabulary_id AND n.status = 1 GROUP BY d.tid, d.name ORDER BY updated DESC, d.name");
  $items = array();
  while ($category = db_fetch_object($result)) {
    $items[] = l($category->name .' ('. $category->count .')', 'taxonomy/term/'. $category->tid) .'<br />';
  }
print theme('item_list', $items); 

Using the above snippet I am able to acheive:

- Holiday (4)
- Work (6)

I want to display:

# Holiday (4) nodes and (20) comments (This means that the term 'holiday' has 4 nodes tagged with the term and the 4 nodes in total have 20 comments)
# Work (6) nodes and (36) comments

Can somebody please help.

Thanks,
Manoj