Weblabor

WP-ben egy oldal sablonmintát szeretnék módosítani, úgy hogy ne az összeset kérdezze le, hanem csak az általam kiválasztottakat, pl.: 5-10 elemet. Az alapprobléma, hogy az oldalon az étlapnál és az itallapnál is megjelenik minden elem (ételek, italok is).

page.template.menu.php

[colorer=php]<?php
/*
* Template Name: Menu Page
*/
get_header(); ?>



<?php //$temp_query = $wp_query; ?>

<!-- begin main -->
<section class="main menulist block grid4 clearfix">





<?php if (have_posts()) : while (have_posts()) : the_post(); ?>

<header>
<h2 class="post-title"><?php the_title(); ?></h2>
</header>

<div class="post-content">
<?php the_content() ?>
</div>

<?php endwhile; endif;?>




<?php

// get initial categories
$categories = get_terms( 'menus', array( 'orderby' => 'name', 'order' => 'ASC', 'hierarchical' => true, 'parent'=>0, 'hide_empty' => false, ) );


foreach ( $categories as $category ) {

// we don't want child categories now, and since get_categories does not support 'depth' parameter, we use ->parent check
if ( $category->parent > 0 ) {
continue;
}

$i = 0;
echo '<h2 class="menu-title level-1">' . $category->name . '</h2>';
query_posts(
array(
'menus' => $category->slug,
'post_type'=> 'menu-item',
//'orderby'=> 'menu_order',
'posts_per_page' => -1
)
);


if ( have_posts() ) : while (have_posts()): the_post(); global $post;

// let's make sure that the post is not also in any of child categories, if it is, skip it ( we don't want to display it twice )
$child_categories = get_term_children( $category->term_id, 'menus' );

//if ( $child_categories && in_category( $child_categories, $post->ID ) ) {
if ( $child_categories && !is_object_in_term( $post->ID, 'menus', $category->name ) ) {
continue;
}

echo 0 === $i ?