Where do I throw the exception? - php

Question

I have the following php function that renders an array of templates based on an array of paths. In other words, if you provide a set of arrays such as:
$array_template = array(
'carousel' => 'carousel', //type=>name (with out extension).
'mini' => 'mini_feed'
)
$array_paths = array(
'path_one' => 'path/to/one/',
'path_two' => 'path/to/two/'
)
To this function:
protected function _render_templates_array($templates, array $template_name){
foreach($template_name as $type=>$name){
foreach($templates as $template=>$path){
if(file_exists($path . $name . '.phtml')){
require_once($path . $name . '.phtml');
}
}
}
return;
}
It should find and render every file, checking every path for that file.
The problem I am having is, I figured out how to make it stop searching once all files are found, how ever, do I append an else to the if and throw my error? or is there some where else I should throw my error?
Essentially I need:
render all templates, making sure to look in all paths for those templates.
throw an error if the template is not found in any path.
stop processing once all files are loaded.
Thoughts?

Answer 1

Between the two foreach lines, add $found = false;. Inside the if, add $found = true; Between the two "end foreach" }, add if(!$found) throw.....; as needed.
protected function _render_templates_array($templates, array $template_name){
foreach($template_name as $type=>$name){
$found = false;
foreach($templates as $template=>$path){
if(file_exists($path . $name . '.phtml')){
require_once($path . $name . '.phtml');
$found = true;
}
}
if( !$found) throw new .......;
}
return;
}