<?php echo get_the_term_list( $post->ID, \'people\', \'People: \', \' \', \'\' ); ?>
People: <a href="person1">Person1</a>, <a href="person2">Person2</a>, ...
People: Person1, Person2
如何从“Get Term List”函数中删除链接？
4 个回复
最合适的回答,由SO网友:Joe Hoyle 整理而成
手动编写列表可能更容易，例如：
<?php
$terms = wp_get_post_tags( $post->ID );
//For custom taxonomy use this line below
//$terms = wp_get_object_terms( $post->ID, \'people\' );
foreach( $terms as $term )
$term_names[] = $term->name;
echo implode( \', \', $term_names );
SO网友:Puneet
或者，您也可以使用
<?php
echo strip_tags (
get_the_term_list( get_the_ID(), \'tax_name\', "Text Before Value ",", " )
);
?>
SO网友:Carson
我找到了另一种更直接回答我问题的方法：
<?php $terms_as_text = get_the_term_list( $post->ID,\'people\', \'People: \', \', \');
if (!empty($terms_as_text)) echo \'<p>\', strip_tags($terms_as_text) ,\'</p>\'; ?>
SO网友:Danny Hefer
如果要将术语显示为HTML列表，则使用strip\\u tags（）可能会变得复杂。这儿，有件东西给你$将raw设置为true（或任何非空的内容）只会创建一个带有您选择的$分隔符的内联列表，如果不是，它将生成一个没有链接的HTML列表。如果希望列表具有样式化的标题，请将$titletag设置为H1或H2。如果不需要标题，只需将$title留空即可。
function show_tax($taxname, $title, $title_tag, $raw, $separator){
$terms = get_the_terms($post->ID, $taxname);
$out = \'\';
if (!empty($title)){
if(empty($title_tag)){
$title_tag = \'span\';
}
$out .= \'<\'.$title_tag.\'>\'.$title.\'</\'.$title_tag.\'>\';
}
if (!empty($raw)){
$out = implode($separator, $terms);
}
else{
$out .= \'<ul>\';
foreach ( $terms as $term ){
$out .=\'<li>\'.$term->name.\'</li> \';
}
$out .= \'</ul>\';
}
return $out;
}
echo show_tax(\'people\', \'PEOPLE\', \'h3\', \'\', \'\'); // An html list with PEOPLE as title
echo show_tax(\'people\', \'PEOPLE:\', \'\', true, \',\'); // An inline list with PEOPLE: as before text
结束