Snippet code to check if current selected menu has active child

Hello,

I was searching the forum for few hours but couldn’t find the solution for my problem..

I need a snippet code that will check if current selected menu has active child, if it has, I need to return menu name, if doesn’t, I have to return empty string.

Thank you!

now the code I have is:

<?php
$id = isset($id)? $id: $modx->documentObject[’id’];
$field = isset($field)? $field : "pagetitle";
$result = $modx->getDocument($id,$field);
return $result[$field];
?>

It returns all the page titles.

http://wiki.modxcms.com/index.php/API:getActiveChildren
http://wiki.modxcms.com/index.php/API:getAllChildren

Solved with the next code:

<?php
$id = isset($id)? $id: $modx->documentObject[’id’];
$field = isset($field)? $field : "pagetitle";
$result = $modx->getDocument($id,$field);
$resu = $modx->getActiveChildren2($id);
if(count($resu))
{
return $result[$field];
}
else
{
return ’’;
}
?>


and this code add style display none inside the div

<?php
$id = isset($id)? $id: $modx->documentObject[’id’];
$field = isset($field)? $field : "pagetitle";
$result = $modx->getDocument($id,$field);
$resu = $modx->getActiveChildren2($id);
if(count($resu))
{
return ’’;
}
else
{
return ’style="display:none;"’;
}
?>

It helps if you use the forum’s code tags to enclose code; it’s the # button.