我对编程很陌生，不知道该怎么做。下面的代码返回3个类别名称和功能图像，然后将它们输出到页面上。

<div id="footer">
    <div id="footer-inside">
        <div id="featured-categories-footer">
            <?php
                $featured_categories = array(8,9,10);
                foreach  ($featured_categories as $featured) {
                    if (get_option(THEME_PREFIX . \'enable_category_\' . $featured)) {
                        echo \'<!-- featured category #\'.$count.\' -->\';
                        $category_id = get_option(THEME_PREFIX . \'featured_category_\' . $featured); 
                        query_posts("cat=$category_id&showposts=1"); if (have_posts()) : while (have_posts()) : the_post(); ?>
                            <div class="featured-category-footer">
                                <a href="<?php echo get_category_link($category_id); ?>" title="View all posts in <?php echo get_cat_name($category_id); ?>">
                                    <?php the_post_thumbnail(); ?>
                                </a>

                                <h3><?php echo get_cat_name($category_id); ?></h3>
                                <?php echo category_description($category_id); ?>
                            </div>  
                        <?php
                        endwhile; endif; 
                        wp_reset_query();
                    }
                }
            ?>
        </div>

我想做的是，我想指向三个静态页面，而不是返回在管理区域中指定的3个类别。因此，它将返回这3个页面的功能图像以及名称。有人能帮忙吗？

我认为需要更改的行是“$featured\\u categories=数组（8,9,10）；”而是指向页面。我走对了吗？

<?php     
   $featured_categories = array(8,9,10);

  wp_reset_query();
                    }
                }
            ?>

<div class="featured-category-footer">
    <a href="link to page">link</a>
     <h3>title</h3>
</div>