列出当前父级的子页面，并带有特色图像和描述

Question

有人可以告诉我如何列出当前页面的子页面以及如何添加特色图像和菜单说明吗？到目前为止，这是我找到的内容，但是我无法弄清楚如何使这些说明起作用。

    <ul>
    <?php $child_pages = $wpdb->get_results("SELECT * FROM $wpdb->posts WHERE post_parent = ".$post->ID." AND post_type = 'page' ORDER BY menu_order", 'OBJECT');    ?>
    <?php if ( $child_pages ) : foreach ( $child_pages as $pageChild ) : setup_postdata( $pageChild ); ?>
    <li>
    <a href="<?php echo  get_permalink($pageChild->ID); ?>" rel="bookmark" title="<?php echo $pageChild->post_title; ?>">
    <span class="thumbnail"><?php echo get_the_post_thumbnail($pageChild->ID, 'small-thumb'); ?></span>
    <span class="title"><?php echo $pageChild->post_title; ?></span>
    </a>
    </li>
    <?php endforeach; endif; ?>
    </ul>

谢谢。

Answer 1

我最终仅使用自定义字段而不是菜单描述。这是对我有用的最终代码。

  <?php 
  $args = array(
      'parent' => $post->ID,
      'post_type' => 'page',
      'post_status' => 'publish'
  ); 
  $pages = get_pages($args);  ?>
  <ul class="four no-bullets"> 
  <?php foreach( $pages as $page ) { ?>
   <li>
      <a href="<?php echo  get_permalink($page->ID); ?>" rel="bookmark" title="<?php echo $page->post_title; ?>">
      <span class="thumbnail"><?php echo get_the_post_thumbnail($page->ID, 'small-thumb'); ?></span>
      <span class="title"><?php echo $page->post_title; ?></span>
      <span class="desc"><?php echo get_post_meta($page->ID, 'desc', true); ?></span>
      </a>
  </li>
  <?php } ?>
  </ul>