Wordpress Taxonomy Title Output

Question

I am trying to display the title of a taxonomy page when viewing it.

So if I were on the tag page, I would go:

<?php if (is_tag()) {?> <h1><?php single_cat_title(); ?></h1> <?php } ?> 

But how would you achieve the same if the user views a particular taxonomy page?

is_taxonomy() does not exist.

Answer 1

Found this answer for someone else.

Follow this guide: http://justintadlock.com/archives/2009/06/04/using-custom-taxonomies-to-create-a-movie-database

Right down, you need a line:

 <?php $term = get_term_by( 'slug', get_query_var( 'term' ), get_query_var( 'taxonomy' ) ); echo $term->name; ?> 

Answer 2

can do superdupereasy with:

 <?php echo get_queried_object()->name; //output $taxonomy->$tax the simple way ?> 

name if a taxonomy term will be used on any page of a custom taxonomy. on something like example.com/books/Fiction , it will be an echo of fiction .

Instead of name you can also use taxonomy , which will be an echo of books .

And in short they say that taxonomies are NOT categories or pages or comments, this is what you say, for example, “products” or “books”. Using this, you unlock the functionality of Post Post Type WordPress. You can create your own archive-products.php , page-products.php , single-products.php and enjoy it. :)

Answer 3

Here is a complete example that changes the title only for taxonomy listing pages using the is_tax() function

 <title><?php global $page, $paged, $post; if (is_tax()) { $term = get_term_by( 'slug', get_query_var( 'term' ), get_query_var( 'taxonomy' ) ); $term_title = $term->name; echo "$term_title | "; } else { wp_title( '|', true, 'right' ); } // Add the blog name. bloginfo( 'name' ); // Add the blog description for the home/front page. $site_description = get_bloginfo( 'description', 'display' ); if ( $site_description && ( is_home() || is_front_page() ) ) echo " | $site_description"; // Add a page number if necessary: if ( $paged >= 2 || $page >= 2 ) echo ' | ' . sprintf( __( 'Page %s', 'twentyten' ), max( $paged, $page ) ); ?></title> 

Answer 4

You can use single_term_title () .

Answer 5

I think that just using the single_cat_title () function is enough.

See here for more details .