set models based on condition in cakephp queries - php

Question

This is probably very easy to do, but I can't seem to get my head around it right now. Let's say in a component in a cakephp application, I have a variable my_model, which contains the model of the corresponding controller that is currently using the component like:
function TestComponent extend Object
{
var $my_model; // can be either User, or Person
function test()
{
$myModelTemp = $this->my_model;
$model = $myModelTemp != 'User' ? $myModelTemp.'->User' : 'User';
$this->$model->find('all');
}
}
As you can see above in my function test() what I'm trying to do is call the correct model based on the value of my_model. So based on the condition, my query will be either:
$this->Person->User->find('all');
Or
$this->User->find('all');
When I do it like I did above, I get an error saying Fatal error: Call to a member function find() on a non-object. In order words, that error means Person->User is not an object (so, it is considered as a string).

Answer 1

What you're saying could be true, however, it can refer to any part of the call.
So either Person or User could be invalid, or together they causes the error. Hard to say.
Try dumping the individual objects using var_dump();
So try:
<?php
echo "<pre>";
var_dump(is_object($this->Person));
var_dump(is_object($this->User));
echo "</pre>";
?>
to determine where you're code goes wrong.
To be clear, that return value needs to be true for it to be an object.
The one that returns false is the likely culprit.
Should your question refer to the correct way to reference an object, an object is basically an array. For example:
<?php
$obj = (object) array("this", "my_function");
?>
The above example casts the array as an object. However, using multiple layers might prove to be more difficult than you'd expect.

Answer 2

Generally, it looks like you might be going about this all wrong. Obviously you want the models to be dynamic, but then you're hard-coding things which defeats the whole point of it being dynamic in the first place.
It also seems like you might be violating the principals of CakePHP and MVC by doing all this in a component. I'm not sure this component should really be manipulating models or assuming which models are currently in use.
However, if you want to evaluate a string as an actual object, you can wrap it in { ... } (this is valid standard PHP syntax, not Cake-specific code).
Try this:
$modelName = $this->my_model;
$model = ($modelName != 'User') ? $this->{$modelName}->User : $this->User;
$model->find('all');
Now, if this doesn't work or you get an error saying it can't find the model(s) you need to ensure the models are actually loaded and initialised in the current scope.