Calling a class's constant in another class's variable

Question

I was wondering if there is any possibility in PHP to do following;

<?php

class boo {
 static public $myVariable;

 public function __construct ($variable) {
   self::$myVariable = $variable;
 }
}

class foo {
  public $firstVar;
  public $secondVar;
  public $anotherClass;

 public function __construct($configArray) {
   $this->firstVar = $configArray['firstVal'];
   $this->secondVar= $configArray['secondVar'];
   $this->anotherClass= new boo($configArray['thirdVal']);
 }
}

$classFoo = new foo (array('firstVal'=>'1st Value', 'secondVar'=>'2nd Value', 'thirdVal'=>'Hello World',));

echo $classFoo->anotherClass::$myVariable;
?>

Expected OUTPUT : Hello World

I am getting following error; Parse error: syntax error, unexpected T_PAAMAYIM_NEKUDOTAYIM

I Googled and it is related to colon (double dots) in $classFoo->anotherClass::$myVariable

I wouldn't like to go all the trouble to change my other classes. Is there anyway around this problem?

Thank you for your help in advance.

P.S. I just didn't want to lose few hours on this to find a way around. I already spent yesterday 2.5 hours to change almost whole Jquery because customer wanted a change and today in the morning I was asked to take the changes back because they didn't want to use it (they changed their mind). I am just trying to avoid big changes right now.

netcoder · Accepted Answer · 2011-09-24 21:23:18Z

13

You need to do:

$anotherClass = $classFoo->anotherClass;
echo $anotherClass::$myVariable;

Expanding expressions to class names/objects for static calls/constants is not supported (but expanding variables, as shown above, is).

edited Sep 24, 2011 at 21:23

answered Sep 24, 2011 at 21:15

netcoder

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5 Comments

Revenant Over a year ago

I don't know, why I haven't tried this one (: Thanks a lot for your help. Need to wait a bit more to accept your answer.

Brad Christie Over a year ago

I'm pretty sure that's still not valid, unless it's new to PHP? EDIT: Also, what about using ${...} to eval a get_class result?

netcoder Over a year ago

@Brad Christie: It's valid in 5.3.

netcoder Over a year ago

@Brad Christie: What do you mean "eval a get_class result"?

Brad Christie Over a year ago

@netcoder: When i think static I think to reference the name of the class, not the instance of the class being implemented in a static format (as 5.3 appears to do). I was having trouble getting $anotherClass::$myVariable to work, so instinctively I though to use get_class (basically emulate "boo::$myVariable", where boo would be get_class($classFoo->anotherClass)) but now I see there's no need. Long story short, I was trying to work around a problem that i didn't know didn't exist. ;p

ETech · Accepted Answer · 2014-06-28 14:39:04Z

0

If you do not care about memory and execution speed, this is correct.
It seems that reference would be better:

$classRef = &$classFoo->anotherClass;
echo $classRef;

Works for me.

answered Jun 28, 2014 at 14:39

ETech

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