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Home/ Questions/Q 8739431
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T10:56:11+00:00 2026-06-13T10:56:11+00:00

I want the variable to be parsed within the array so when I echo

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I want the variable to be parsed within the array so when I echo $head[‘meta_title’] , lol is displayed. I have tried wrapping it in double quotes but that doesn’t seem to work either, is there any way round this?? Thanks!

I am getting either unexpected T_VARIABLE and when I use double quotes I get unexpected “”

$meta_title = "lol";

public $head = array
(
    "title"        => "blah",
    "meta_title"   => $meta_title,
    "meta_content" => $meta_content
);
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-13T10:56:12+00:00Added an answer on June 13, 2026 at 10:56 am

    You cannot use an expression to initialize a class property. The values of the two variables are not known until runtime, and therefore can’t be used in the declaration. Instead, define them in the constructor.

    public $head = array
    (
        // The title as a string literal is ok...
        "title"        => "blah",
        "meta_title"   => NULL,
        "meta_content" => NULL
    );
    // Pass them to the constructor as parameters
    public function __construct($meta_title, $meta_content)
    { 
      // Initialize them in the constructor. 
      $this->head['meta_title'] = $meta_title;
      $this->head['meta_content'] = $meta_content;
    }
    

    From the docs

    Class member variables are called “properties”. You may also see them referred to using other terms such as “attributes” or “fields”, but for the purposes of this reference we will use “properties”. They are defined by using one of the keywords public, protected, or private, followed by a normal variable declaration. This declaration may include an initialization, but this initialization must be a constant value–that is, it must be able to be evaluated at compile time and must not depend on run-time information in order to be evaluated.

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