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View Full Version : quicksearch member in $show array


helltonic
10-03-2009, 09:23 PM
Hello, could somebody please tell me the criteria for when 'quicksearch' would be a member of $show? Is it just when the user is logged in? Or is there some other criteria that has to be met?

<a id="navbar_search" href="/forums/search.php$session[sessionurl_q]">Search</a>
<if condition="$show['quicksearch']">
<script type="text/javascript">
vbmenu_register("navbar_search", 1);
</script>
</if>


If I try to see all the members of $show in PHP, it does not include 'quicksearch':

echo('[');
print_r($show);
echo(']');


Prints:

[Array ( [search_engine] => [old_explorer] => [left_column] => 1 [center_column] => 1 [right_column] => 1 [editor_css] => [xfire] => )]

Lynne
10-03-2009, 09:31 PM
The $show array is just full of variables which are either true or false. In the vbcode somewhere, php is used to determine whether something should be shown. If it should, it is set to true. If not, it is set to false. In regards to quicksearch, somewhere (do a search in the files for $show['quicksearch'] ) there is some code to determine whether quick search should be shown and it is set to true or false based on that code.