有人可以告诉我如何列出当前页面的子页面并包含精选图片和菜单说明吗?这是我到目前为止所发现的,但我无法弄清楚如何使描述起作用。
<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>
感谢。
答案 0 :(得分:4)
我最后只使用自定义字段而不是菜单说明。这是最适合我的代码。
<?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>
答案 1 :(得分:2)
请尝试以下代码:
<?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>