PHP Variable re-witting is it possible?

jerry_louise

Is it possible to rewrite

$this->Router->Template->{variable name}

to sometime more readable

$@{variable name}

So to make code more readable when sending variables from my controllers to views in a way that distinguishes them via the @ symbol or similar.

bradgrafelman

jerry_louise wrote:
to make code more readable

Er... huh? You think this:

$@{variable name}

is more readable than this:

$this->Router->Template->{variable name}

??? I think you've got it backwards. 😉

You could always store a reference to $this->Router->Template in a variable with a short name, but I still think you should stick with using the full path to the desired variable/property - it keeps your code more readable. If you come back a week or two after not having worked with your code in order to make some modification or enhancement, you'd probably be able to understand $this->Router->Template->{variable name} a lot easier than $@{variable name} .

jerry_louise

Well for variables that are passed to views.

$this->Cabbit->Template->myVar becomes $myVar in the view.
I would rather view variables were more like $@myVar to $@myVar or something similar to keep them nice and simple, writing out the hold $this->Cabbit->Template->xxx is so long and tedious when in Ruby on Rails i am used to writing @myVar for variables that belong to views.

Is it possible to reference $this->Cabbit->Template-> as something else but keeping it so that the individual variable can be unique?

NogDog

Are you using a specific framework? For example, I know that with CodeIgniter you can pass an array to the view, which will then extract() that array into a set of variables which can then be used directly in the view, e.g.:

controller:

$data = array(
   'fubar' => $this->foo->bar->method(),
   'var' = $this->another->object->method()
);
$this->load->view('view_name', $data);

View:

<h3><?php echo $fubar; ?></h3>
<p><?php echo $var; ?></p>

Weedpacket

The symfony framework also has a mechanism for this. In the controller you write

$this->fubar = "whatever";

and in the view

echo $fubar

Generally, what you're after is an MVC framework: you'd probably prefer one that takes the inspiration for its design from Rails. Cake and symfony are the first that spring to mind for me, NogDog has suggested CodeIgniter, and I recall a project with a name like "PHP on Tracks" or something that's obviously intended to be Rails ported to PHP; there are undoubtedly others out there, and one of them may be more comfortable for you.

That said; for the programmer there's always the pastime of writing your own 🙂...

jerry_louise

Weedpacket;10955344 wrote:
The symfony framework also has a mechanism for this. In the controller you write
$this->fubar = "whatever";
and in the view
echo $fubar
.

Generally, what you're after is an MVC framework: you'd probably prefer one that takes the inspiration for its design from Rails. Cake and symfony are the first that spring to mind for me, NogDog has suggested CodeIgniter, and I recall a project with a name like "PHP on Tracks" or something that's obviously intended to be Rails ported to PHP; there are undoubtedly others out there, and one of them may be more comfortable for you.

That said; for the programmer there's always the pastime of writing your own 🙂...

I am writing my own framework thats why i am trying to work this out, my variables for passing onto views are way to long.

NogDog

I guess it then depends on how you are accessing/running the view, but it should be fairly easy to implement something where you create an associative array, then extract() that array in the view in order to have a set of variables that are named as per that array's keys. Or for that matter, if you are simply including/requiring the view, then just creating whatever variable names you want before the include should be sufficient.

class SomeController extends Controller
{
   public function myMethod()
   {
      $foo = $this->some->method();
      require 'path/to/view.php';
   }
}

view.php:

<?php echo $foo; ?>