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Trond Åge Kvalø
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Monday 02 February 2004 5:02:09 am
Sorry if this is question is too basic, but I've spent several hours searching the forums and docs now without success. So here goes: I have a page with one top menu and one left menu.
In the left menu I'd like to have all folders that belongs to section 9.
The top menu should consist of all folders that does NOT belong to section 9. Any ideas how I can achieve that?
Best regards
Trond Kvalo
Senior developer
Atlantia as
Norway
trondåge
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Claus Jensen
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Monday 02 February 2004 5:34:12 am
Hey there,
You could do this with fetches and {section show=eq($section_id, 9)} ...blablabla.. etc and {/section}
fetch could for instance be:
{section loop=fetch( content, list, hash( section_id, 9,
class_filter_type, include,
class_filter_array, array( 'folder' ),
sort_by, $node.sort_array }
{$:item.name}
{/section}
this may not be totally correct but something like that should work, better check the correct syntax for fetching with section_id's in the docs.
Cheers,
claÜs
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Kevin Myles
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Friday 23 July 2004 9:57:43 am
I was hoping to do the same and I found a different thread ultimately led to my success.
read the last post by Jan http://www.ez.no/community/forum/developer/fetch_content_by_section_id basically, it tells you to add an attribute_filter which of course, is the section, However, I found that you had to replace the $node_section_id or whatever it was) with the section you wanted, so in my case, the end result code was: {let menuitems=fetch( 'content', 'list', hash( 'parent_node_id', 2,
'sort_by', $root_node.sort_array, attribute_filter, array( 'and', array( 'section', '=', 1 ) ),
class_filter_type, include,
class_filter_array, array( 'folder' ) ) )}
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