variable variables + objects in fields

Question

I am trying to do something with variable variables and I got stuck on an object problem. Imagine this class setup:

class A
{
  public $field = 10;
}

class B
{
  public $a;

  public function __construct()
  {
    $this->a = new A();
  }
}

Now everyone knows that this pice of code works:

$a = new A();
$var = 'field';
echo $a->$var; // this will echo 10

Is there any possibility I could make something like this work?:

$b = new B();
$var = 'a->field';
echo $b->$var; // this fails

Note: any option which does not use eval function?

Well it doesnt work :) That is the question, how to make it work...how to write the code so something like that works if it is even possible — Santhos
– Santhos, Commented Apr 18, 2012 at 19:12

Dmytro Shevchenko · Accepted Answer · 2012-04-18 19:27:48Z

2

How about using a closure?

$getAField = function($b) {
    return $b->a->field;
};

$b = new B();
echo $getAField($b);

Though, it's only possible in newer versions of PHP.

Or, as a more generic version, something like this:

function getInnerField($b, $path) { // $path is an array representing chain of member names
    foreach($path as $v)
        $b = $b->$v;
    return $b;
}

$b = new B();
echo getInnerField($b, array("a", "field"));

edited Apr 18, 2012 at 19:27

answered Apr 18, 2012 at 19:15

Dmytro Shevchenko

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Comments

Benjamin Schulte · Accepted Answer · 2012-04-18 19:25:22Z

1

You can write a custom __get method on your class to access the childrens property. This works:

class A
{
  public $field = 10;
}

class B
{
  public $a;

  public function __construct()
  {
    $this->a = new A();
  }

  public function __get($property) {
    $scope = $this;

    foreach (explode('->', $property) as $child) {
      if (isset($scope->$child)) {
    $scope = $scope->$child;
      } else {
    throw new Exception('Property ' . $property . ' is not a property of this object');
      }
    }

    return $scope;
  }
}

$b = new B();
$var = 'a->field';
echo $b->$var;

Hope that helps

answered Apr 18, 2012 at 19:25

Benjamin Schulte

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Comments

Leven · Accepted Answer · 2012-04-18 19:13:45Z

0

I don't recommend it, but you could use eval:

$b = new B();
$var = 'a->field';
eval( 'echo $b->$'.$var );

answered Apr 18, 2012 at 19:13

Leven

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Comments

Leven · Accepted Answer · 2012-04-18 19:23:07Z

0

This should also work I guess:

$b = new B();
$var1 = 'a'; 
$var2 = 'field'

echo ($b->$var1)->$var2;

answered Apr 18, 2012 at 19:23

Leven

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