Saturday 29 March 2003 7:23:21 pm - 4 replies
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}
Sunday 30 March 2003 5:19:33 am
Well i found the solution by myself.
1º the script doesn't worked because with ViewCaching=true the variable $node doesn't exist
2º i can get section_id in variable $module_result, so i don't need $node anymore
3º i found a better way to implement like this:
{let children=fetch('content','list',hash(parent_node_id,40))}
{section name=Child loop=$children}
<div class="menu{cond($module_result.section_id|eq($Child:item.object.section_id),"on","off"
}">
<a href={$Child:item.url_alias|ezurl}>{$Child:item.name}</a>
</div>
{/section}
Regards
Paulo Almeida
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