PHP Reference variable in a class

Question

I am trying to reference a public variable from inside a class.

class Settings{
    public $CompanyName = "MyWebsite"; // company name
    public $PageTitle   = "$this->CompanyName - big website"; // E.g. My Big Website
}

But this gives back a parse error:

Parse error: parse error

What is the correct way of doing this?

`"$this->CompanyName - big website"` should be `$this->CompanyName ." - big website"` — Dave, Apr 30 '13 at 12:41
@Dave - No it shouldn't, that'll still produce a parse error, because you can't initialize member variables to anything that is non-static. — nickb, Apr 30 '13 at 12:42
Yeah I just fixed the obvious error didn't even look at the rest tbh — Dave, Apr 30 '13 at 12:44

score 3 · Answer 1 · answered Apr 30 '13 at 12:46

you can't use $this on the properties, but in the methods, try to define the pagetitle in the __construct();

class Settings{
    public $CompanyName = "MyWebsite"; // company name
    public $PageTitle;

    function __construct(){
        $this->PageTitle = "$this->CompanyName - big Website";
    }
}

$settings = new Settings();
echo $settings->PageTitle;

outputs: MyWebsite - big Website

score 2 · Accepted Answer · answered Apr 30 '13 at 12:45

You can't set variable with other variable when defining. Use __construct for it:

class Settings{
    public $CompanyName = "MyWebsite"; // company name
    public $PageTitle; // E.g. My Big Website

    public function __construct(){
        $this->PageTitle = $this->CompanyName." - big website";
    }
}

egig · Answer 3 · 2013-04-30T12:56:54.473

http://php.net/manual/en/language.oop5.properties.php :

This declaration may include an initialization, but this initialization must be a constant value

it's not valid.

this is not valid:

public $var1 = 'hello ' . 'world';

but this is:

public $var1 = myConstant;
public $params = array();

I'll do it this way :

class Settings{
    public $CompanyName;
    public $PageTitle; // E.g. My Big Website

    public function __construct(){
       $this->$CompanyName = 'mywebsite';
       $this->PageTitle = $this->CompanyName." - big website";
   }
}

PHP Reference variable in a class

3 Answers3