Fetching Taxonomy Slug - php

Question

I am using the following code to get taxonomy slug:
<?php
$terms = get_the_terms( $post->ID, 'locations' );
if ( !empty( $terms ) ){
$term = array_shift( $terms );
}
?>
I am then using the following code to output the slug:
<?php echo $term->slug; ?>
My question is, how can I use this to output two different taxonomies in the same location? For example:
<?php
$terms = get_the_terms( $post->ID, 'locations', 'status' );
if ( !empty( $terms ) ){
$term = array_shift( $terms );
}
?>
I thought I could perhaps add terms 'location' , 'status' but it doesn't work.

Answer 1

If you want to display two or more taxonomy then i think you should loop the $terms variable.
<?php
$terms = get_the_terms( $post->ID, 'locations' );
if ( !empty( $terms ) ){
foreach ($terms as $term):
echo $term->slug;
endforeach;
}
?>
Hope its helps you.
Thank You

Answer 2

According to the official documentation for get_the_terms, only one taxonomy can be supplied. If you wanted to output the slugs of all the terms within two different taxonomies, you can do as Mohammad suggested, but twice.
i.e.
<?php
// output all slugs for the locations taxonomy
$locations_terms = get_the_terms( $post->ID, 'locations' );
if ( ! empty( $locations_terms ) ) {
foreach ( $locations_terms as $term ) {
echo $term->slug;
}
}
// output all slugs for the status taxonomy
$status_terms = get_the_terms( $post->ID, 'status' );
if ( ! empty( $status_terms ) ) {
foreach ( $status_terms as $term ) {
echo $term->slug;
}
}
?>
However, if you only care to get the slug of an individual term in each of the taxonomies, you may find get_term_by more useful.
i.e.
<?php
$loc_field = 'name';
$loc_field_value = 'special location';
$loc_taxonomy = 'locations';
$locations_term = get_term_by( $loc_field, $loc_field_value, $loc_taxonomy );
echo $locations_term->slug;
$stat_field = 'name';
$stat_field_value = 'special status';
$stat_taxonomy = 'status';
$status_term = get_term_by( $stat_field, $stat_field_value, $stat_taxonomy );
echo $status_term->slug;
?>