PHP Use Include Inside Function

Question

Im trying to make a function that I can call as follows,

view( 'archive', 'post.php' );

and what the function really does is this.

include( 'view/archive/post.php' );

The reason for this is if in the future I expand the directory to be view/archive/another_level/post.php I dont want to have to go back everywhere in my code and change all the include paths.

Currently this is what i have for my function, except it appears that the include is being call inside the function, and not being called when the function is called...

function view( $view, $file )
    {
        switch ( $view )
        {
            case 'archive'  : $view = 'archive/temp';   break;
            case 'single'   : $view = 'single';     break;
        }

        include( TEMPLATEPATH . "/view/{$view}/{$file}" );
    }

How can I get this function to properly include the file?

EDIT:

There were no errors being displayed. Thanks to @Ramesh for the error checking code, ini_set('display_errors','On') I was able to see that there were other 'un-displayed' errors on the included file, which appeared to have caused the file not to show up...

Answer 1

Here's one way you could solve the problem:

change the way you call your function so it looks like this:

include( view('archive','post') );

and your function would look like this:

<?php
function view( $view, $file )
{
    switch ( $view )
    {
        case 'archive': $view = 'archive/temp'; break;
        case 'single' : $view = 'single';       break;
    }

    return TEMPLATEPATH . "/view/{$view}/{$file}";
}
?>