Best way of creating Class in a snippet

40 Posts

Benjamin Abel Reply #1, 10 years, 7 months ago

Lets say I have some Controller class in a snippet:

class Controller {...}
$ctrl = new Controller();

At the bottom of this snippet, I have a switch which controls what method I need to launch:

switch($scriptProperties['action']) {
    case 'do':
        return $ctrl->doSomething();
        break;
    ...
}

The thing is, this snippet must be called multiple times on a page. Which means I have to check if such class doesn't already exists (not a big problem, thou) AND creating a new instance of this class each time.

Is there any better way to do such things? Thanks in advance.

This question has been answered by roman.khazizov. See the first response.

24,544 Posts

BobRay Reply #2, 10 years, 7 months ago

The best option is to make doSomething() a static method and call it with:

return Controller::doSomething();

Then there's no need to instantiate the class.

This may not be possible, depending on what the controller does, because you can't use $this to access class variables in a static method.

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40 Posts

Benjamin Abel Reply #3, 10 years, 7 months ago

Quote from: BobRay at Mar 03, 2016, 02:48 PM

This may not be possible, depending on what the controller does, because you can't use $this to access class variables in a static method.

That's why I ended up using singleton pattern. This might actually solve the problem:

public static function getInstance()
    {
        if (null === static::$instance) {
            static::$instance = new static();
        }
        
        return static::$instance;
    }

This way it should instantiate a class only once.

24,544 Posts

BobRay Reply #4, 10 years, 7 months ago

That looks like a good solution.

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Answered Best way of creating Class in a snippet