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Home/ Questions/Q 6248791
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Editorial Team
  • 0
Editorial Team
Asked: May 24, 20262026-05-24T13:07:23+00:00 2026-05-24T13:07:23+00:00

I couldn’t find a answer to my question and I am thinking there is

  • 0

I couldn’t find a answer to my question and I am thinking there is something easy I am missing..

I am trying to reference a a value within a object with a variable in a class. In this case I want the line on the bottom:

echo $b->ref->$a->type

to output ‘testing’ like the following two will:

echo $b->ref->test; // outputs 'testing'
$c = $a->type;
echo $b->ref->$c; // outputs 'testing'

Full code:

    <?php

class A {

    public $type;

    public function set_type($type) {
        $this->type = $type;
    }
}

class B {
    public $ref;

    public function set_reference($ref) {
        $this->ref = $ref;
    }

}

$a = new A();
$b = new B();
$b->set_reference( (object) array('test' => 'testing', 'test2' => 'testing2') );

$a->set_type('test');

echo $b->ref->test; // outputs 'testing'
echo '<br />';
echo $a->type; // outputs 'test'
echo '<br />';

$c = $a->type;

echo $b->ref->$c; // outputs 'testing'
echo '<br />';

echo $b->ref->$a->type; // error

What am I missing to be able to do this? Or, is this not possible?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-24T13:07:24+00:00Added an answer on May 24, 2026 at 1:07 pm

    Same as always.

    echo $b->ref->{$a->type};
    
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