我返回了来自三种不同分类法的术语列表。

$terms = get_terms(array("air-categories","web-categories","ink-categories"));

我得到了这个列表，我需要根据它们所处的分类法将它们链接到我网站的不同部分，但我似乎想不出一种方法来返回if语句，每个术语所处的分类法。

global $post;
$published_posts = wp_count_posts()->publish;
$terms = get_terms(array("air-categories","web-categories","ink-categories"));
    $count = count($terms);
    if ( $count > 0 ){
        echo "<ul>";
        foreach ( $terms as $term )             {
            $c = $term->count;
            $term_percent = log($c)*5;
                                                    echo "<li><a href=\'";
        if( ***would like to check for the taxonomy here*** ) {
              echo "***and then change the link here based on the result***";
         }
        echo "\'><span class=\'label\'>" . $term->name . "</span><span class=\'bar\' style=\'width:".$term_percent."%;\'></span><span class=\'count\'>". $term->count ."</span></a></li>";
        }
        echo "</ul>";
                                                 }
    wp_reset_query();

$terms = get_terms( array( \'category\', \'post_tag\' ) );

foreach ( $terms as $term )
{
    $tax_name = esc_html( get_taxonomy( $term->taxonomy )->labels->name );
    echo "$term->name ($tax_name)<br>";
}