列出當前父母的子頁面與特色圖像和說明

Question

有人可以告訴我如何列出當前頁面的子頁面，幷包括精選圖像和菜單說明？這是我迄今爲止發現的，但我無法弄清楚如何讓描述工作。

<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> 

謝謝。

Answer 1

最後我只使用一個自定義字段，而不是菜單描述。這是我工作的最終代碼。

<?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> 

Answer 2

嘗試下面的代碼：

<?php 
    $args = array(
     'parent' => 2, 
     'post_type' => 'page', 
     'post_status' => 'publish' 
    ); 
    $pages = get_pages($args); ?> 
    <ul> 
    <?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> 
     </a><span class="desc"><?php echo $page->post_content; ?></span> 
    </li> 
    <?php 
    } 
    ?> 
    </ul>