Different treemenu for different folder

Friday 17 February 2006 2:31:06 pm

Hi

I have the following structure of content:

-folder (node: 50)
.|
.|-folder1 (node: 60)
...|
...|-article
...|-article
...|-folder
.......|-article
.......|-article
...|
...|-folder
.......|-article
.......|-article
.|
.|-folder2 (node: 70)
...|
...|-article
...|-article
...|-folder
.......|-article
.......|-article

Depends on which item I click, I would like to have a list like a menu.

I could set up an override file for each folder (70 and 60) and make a list with this code:

{let category_list=fetch( content, list, hash( parent_node_id, [NODE_FOLDER],
                                               class_filter_type, include,
                                               class_filter_array, array('folder','article') ) )}

   {section var=category loop=$category_list}
      <a href={$category.item.url_alias|ezurl}>{$category.item.name|wash}</a>
                           
         {let news_list=fetch( content, list, hash( parent_node_id, $category.item.node_id,
                                                    class_filter_type, include,
                                                    class_filter_array, array('article') ) )}

            {section var=newsitem loop=$news_list}
               <a href={$newsitem.item.url_alias|ezurl}>{$newsitem.item.name|wash}</a>
            {/section}

         {/let}
   {/section}
{/let}

I will have many itens inside the folder(50) and I wouldn't like to create an override file for it every item.

How can I use treemenu operator or somenthing like that to make a different menu for each item inside the folder(50) without have to create many override files?

If I click on folder1(60) I would like to have a menu with all its contents and keep this menu while its children were accessed. It would be the same for folder2(70).

Thanks in advance!

Leandro