php – 如何从WP_Post对象获取标题
发布时间:2020-12-13 18:09:59 所属栏目:PHP教程 来源:网络整理
导读:我试图看到子页面列表名称与一些描述显示..i使用下面的代码 $my_wp_query = new WP_Query();$all_wp_pages = $my_wp_query-query(array('post_type' = 'page'));// Get the page as an Object$portfolio = get_page_by_title('service');// Filter through a
我试图看到子页面列表名称与一些描述显示..i使用下面的代码
$my_wp_query = new WP_Query(); $all_wp_pages = $my_wp_query->query(array('post_type' => 'page')); // Get the page as an Object $portfolio = get_page_by_title('service'); // Filter through all pages and find Portfolio's children $portfolio_children = get_page_children( $portfolio->ID,$all_wp_pages ); // echo what we get back from WP to the browser echo "<pre>";print_r( ); foreach($portfolio_children as $pagedet): echo $pagedet['post_title']; endforeach; 我在使用foreach之前得到数组 when i print $portfolio_children iam getting out put like this Array ( [0] => WP_Post Object ( [ID] => 201 [post_title] => Website Hosting ) [1]=> WP_Post Object ( [ID] => 202 [post_title] => Website ) 在foreach之后,如果我打印$pagedet我得到 WP_Post Object ( [ID] => 201 [post_title] => Website Hosting ) 我试着打电话给$pagedet [‘post_title’]但是id没有显示任何东西……提前谢谢
可以肯定的是,您应该将每个页面数据用作列名.
例如, $page_data->post_content //is true,$page_data->the_title // is false. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |