如何在 drupal 7 中列出当前父页面的所有子页面？

Question

Suppose I have parent pages A1 and B1. A1 has child pages A1.1, A1.2,A1.3 and B1 has child pages B1.1, B1.2. I want to list all the respective child pages on A1 and B1. In every child page I have an image and a title. These 2 information needs to be listed in the form of a teaser on the parent page. I need help in doing this whether by coding or by using views, I don't mind as far as I get the proper results. Thank you

Answer 1

You can do this is views by creating a view displaying the fields you require or a teaser. Then add a "Content Nid" contextual filter, in the configeration for this filter under "WHEN THE FILTER VALUE IS NOT AVAILABLE" select "Provide default value" and then "PHP Code" then the code I use is as follows

$children = array();
$current = db_query("select menu_name, mlid from {menu_links} where link_path = :node", array(':node' => $_GET['q']));
$current_info = array();
foreach ($current as $value) {
$current_info[] = $value;
}
if($current_info) {
$result = db_query("select mlid, plid, link_path, link_title from {menu_links}    where menu_name=:menu and plid=:mlid and hidden=0 order by weight, link_title", array(':menu' => $current_info[0]->menu_name, ':mlid' => $current_info[0]->mlid));
foreach ($result as $row) {
  $children[] = $row;
}
}
$nids = array();
foreach ($children as $value){
if( substr( $value->link_path, 0, 5 ) == 'node/' ){
  $nids[] = substr( $value->link_path, 5 );
}
}
return implode('+',$nids);

The last thing to do, under "more" at the bottom of the page sellect "Allow multiple values"

Answer 2

I'm using menu_tree_all_data() to get whole menu structure and then I'm "manually" crawling menu tree.

Also just after reading the tree, I'm calling menu_tree_add_active_path() which will add active trail indicator. It's part of menu block module so you'll have to install it and don't forget to add dependencies for menu block in your module.

  $tree = menu_tree_all_data($menu);
  menu_tree_add_active_path($tree);