PHP 5.3 Dynamically instantiating a class without setting a variable first

Question

OK, this has been a personal bugbear of mine for quite some time. Say I have a class.

class One {
    public $class = 'Two';
    public $member = 'member';
}

class Two {
    public $member = 'Hey there';

    function __construct() {
        print 'Created';
    }
}

$one = new One();

// case 1: works
$two_class = $one->class;
$two = new $two_class();

// case 2: fails
$two = new {$one->class}();

Is there any way to instantiate a class from a class memeber without first assigning the name to a variable? I die a little inside every time I want to create a class dynamically from a property, and I have to populate a variable first. Can anyone explain to me technically why this doesn't work when:

print $two->{$one->method}

Will happily print 'Hey there'?

Technically you don't need the ()'s to start that new class. Also, try assigning it to a variable $new_one = $one->class; $two = new $new_one(); — wesside
– wesside, Commented Aug 28, 2012 at 13:21
I would always discourage you to do anything like this. Simply because you are nto sure whether the given class has the functionality you think it has... It's a maintenance horror! Rather just create a function that creates an object based on the value of the string like function GetMeMyClass($input) { switch ($input) { case 'One': return new One(); case 'Two': return new Two(); default: die("Invalid class given in GetMeMyClass"); } } — bkwint
– bkwint, Commented Aug 28, 2012 at 13:24
@bkwint I wasn't after your personal preference when it comes to OOP. I just wanted a solution to the question posed. Thanks for your thoughts though. — thepearson
– thepearson, Commented Aug 28, 2012 at 13:34

Peter · Accepted Answer · 2012-08-28 13:20:00Z

9

$two = new $one->class();

Demo: http://codepad.org/64iCiWn2

But you gonna get big trouble if $one->class() is function - it may be confusing, but same thing will happen if if $two_class become function

answered Aug 28, 2012 at 13:20

Peter

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