post_status question with wp_list_pages() function

I have a need to build a menu of page links horizontally along my footer element… no issues on design or using wp_list_pages() to do it, etc but I need to have the resulting link(s) not display if that page isn’t published at the time, to not display.

I know normally it wouldn’t if I just used:

<?php wp_list_pages('include=4&title_li='); ?><?php wp_list_pages('include=52&title_li='); ?><?php wp_list_pages('include=17&title_li='); ?>

BUT, I need to have something look selected as well when appropriate — even if on a child page of one of the links… I can hardcode this easily (real example below):

<li <?php if (is_page(4)) {echo ' class="footselected"'; } else {} ?>><span>|</span>&nbsp;&nbsp;<a href="url">Contact Us</a>&nbsp;&nbsp;</li><li<?php if (is_page(52)) {echo ' class="footselected"'; } else {} ?>><span>|</span>&nbsp;&nbsp;<a href="url">Subscribe</a>&nbsp;&nbsp;</li><li<?php if (is_page(17)) {echo ' class="footselected"'; } else {} ?>><span>|</span>&nbsp;&nbsp;<a href="url">Resources</a>&nbsp;&nbsp;</li>

The problem with trying to use wp_list_pages() to generate both the link and text of the link is that A) I cannot use my conditional class=”footselected” code because there’s no place INSIDE the wp_list_pages() function to pass that…. is there?

So, I thought that maybe there was a way to use something equivalent to if post_status == ‘publish’ wrapped around each or the group?

Any hints, folks?

I know I can use it this way:

<?php
// Set output to be empty
$output = '';
// Get pages
$pages = get_pages("child_of=18&sort_column=post_title&title_li=");
// Build output
foreach ($pages as $page) {
	$output .= ($page->post_status == 'publish') ? '<div class="clients">'.$page->post_excerpt.'</div>' : '';
}
// Display output
echo $output;
?>

to generate markup and metadata for a page or child page, but cannot figure how to apply that to the footer links where I only want a specific few pages — not all that are published and I want to use wp_list_pages().