我目前正在 Wordpress 主题中进行修改后的搜索,该主题查询自定义分类location
并将其术语显示为搜索结果。我找不到内置的 Wordpress 函数来处理这个问题,所以我使用了$wpdb query
$keywords = $_GET['s'];
$results = $wpdb->get_results( "SELECT * FROM $wpdb->terms WHERE name LIKE '%%$keywords%%'");
<ul>
<?php foreach ( $results as $result ) :?>
<li><?php echo $result->name;?></li>
<?php endforeach;?>
</ul>
问题在于该表wp_terms
不仅存储自定义分类术语,还存储其他默认术语。因此,为了仅显示自定义分类法的搜索结果,而不是其他默认术语,我想使用get_terms
来获取属于自定义分类法的所有术语,并根据结果location
显示表中的术语wp_terms
get_terms
in_array
$keywords = $_GET['s'];
$results = $wpdb->get_results( "SELECT * FROM $wpdb->terms WHERE name LIKE '%%$keywords%%'");
$terms = get_terms("location");
<ul>
<?php foreach ( $results as $result ) :?>
if(in_array($result->name, $terms)) :?>
<li><?php echo $result->name;?></li>
<?php endif;?>
<?php endforeach;?>
</ul>
但是,$results
and$terms
都是 stdClass 对象,所以in_array
不起作用。
是否有函数、方法或可能的 MySQL 查询允许我根据对象$results
的内容显示对象的结果$terms
?
提前致谢。
编辑:
内容$terms
Array (
[0] => stdClass Object ( [term_id] => 32 [name] => US [slug] => us [term_group] => 0 [term_taxonomy_id] => 32 [taxonomy] => signs [description] => [parent] => 25 [count] => 1 )
[1] => stdClass Object ( [term_id] => 22 [name] => EU [slug] => eu [term_group] => 0 [term_taxonomy_id] => 22 [taxonomy] => signs [description] => [parent] => 0 [count] => 3 )
[2] => stdClass Object ( [term_id] => 26 [name] => AU [slug] => au [term_group] => 0 [term_taxonomy_id] => 26 [taxonomy] => signs [description] => [parent] => 22 [count] => 1 )
[3] => stdClass Object ( [term_id] => 27 [name] => IE [slug] => ie [term_group] => 0 [term_taxonomy_id] => 27 [taxonomy] => signs [description] => [parent] => 22 [count] => 2 )
[4] => stdClass Object ( [term_id] => 23 [name] => PK [slug] => pk [term_group] => 0 [term_taxonomy_id] => 23 [taxonomy] => signs [description] => [parent] => 0 [count] => 2 )
)