Left menu with ViewCaching=true

Saturday 29 March 2003 10:23:21 am

Hi,

My left_menu.tpl (in the bottom), that i include in pagelayout.tpl, creates an menu with diferent <div> if the section of each node is the same of the actually show node. This executes well if ViewCaching=false. If true give same <div> for all nodes.
Can anyone help me? ViewCaching true is much speedy, so i want try to get a solution without turning it to false, and without making a left_menu static, is it possible?

Thank in advance

Paulo Almeida

left_menu.tpl

{let children=fetch('content','list',hash(parent_node_id,40))}

{* loop children and print name with link *}
{section name=Child loop=$children}
{switch name=sw match=$Child:item.object.section_id}

{case match=$node.object.section_id}
<div class="menuon">
{/case}
{case}
<div class="menuoff">
{/case}
{/switch}
<a href={$Child:item.url_alias|ezurl}>{$Child:item.name}</a>
</div>
{/section}

{/let}

PACPI.COM Internet Consulting
http://pacpi.com

Sunday 30 March 2003 11:32:04 pm

Paulo,

You can also use $DesignKeys:used.node To view all DesignKeys use this in the body of your document {$DesignKeys:used|attribute(show, 1)} to show all the variables available.

I hope this helps

Tony

Tony Wood : twitter.com/tonywood
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Thursday 12 June 2003 7:40:32 am

Hello,

I've solved it using a big
{let node=fetch('content','node',hash(node_id,$DesignKeys:used.node)) }

...

{/let}
around my whole pagelayout.tpl

Mads Buus Jensen