[SOLVED] variable as classname

[TiloW]

Hey guys =)

 

Here's my problem:

// exemplary class
class Test
{
  static public DoSomething()
  {}
}

// reference of class "Test"
$testClassName = "Test";

// create new instance of class "Test"
$testInstance = new $testClassName();  // works just fine

// call static method of class "Test"
$testClassName::DoSomething(); // doesn't work

 

So.. How can i call a static method of a class which is strored in a variable?

 

Thanks for any help =)

[.josh]

// reference of class "Test"

$testClassName = "Test";

 

$testClassName is not an object. It's a variable, and you're assigning a regular string to it.

[TiloW]

true.. but in some situations this string is interpreted as a classname - i've provided one example above.

Is there any other way to store classes in variables?

Why does the instantiation work, but not the call of a static method?!

[.josh]

That's not an example of of a string being interpreted as a class name. That's an example of storing a string of value "Test" to a variable $testClassName.  If you want to reference a static method like that, you can do

 

Test::doSomething();

 

or else like

 

$testClassName = "Test";

eval($testClassName."::doSomething();");

[ignace]

You may want to take a look at the factory method design pattern: http://en.wikipedia.org/wiki/Factory_method_pattern

[TiloW]

Well.. I don't really get the difference because in

$testInstance = new $testClassName();

the value of $testClassName is interpreted as a name of a class ..isn't it?

 

Anyway.. the eval function is exactly what i was looking for!

 

thanks Crayon =)

[.josh]

Well.. I don't really get the difference because in

$testInstance = new $testClassName();

the value of $testClassName is interpreted as a name of a class ..isn't it?

 

No.  Class names (and functions) are not parsed inside quotes like that.  Only variables arrays and objects.

[TiloW]

but why does it work? ^^ *confused

[.josh]

Why does what work...this?

Well.. I don't really get the difference because in

$testInstance = new $testClassName();

the value of $testClassName is interpreted as a name of a class ..isn't it?

 

edit:

 

Accidentally hit post.  It does work, because it will create an instance of the object using the value of that variable.  But that is not the same as linking that variable to the class, or somehow making an object out of a variable.  You are simply making use of the value.

[Daniel0]

It works in PHP 5.3.

 

Edit: The static calling with a variable as class name (to prevent ambiguity).

[.josh]

yeah i just edited my post.  I meant to say it doesn't work in the context of it's not the same thing as what he's trying to do as far as using a plain old var as an object.

[TiloW]

okay.. thanks guys