Php, optional public variable
Question by Malixxl
i’ve some public variables in my class, here is how i define them
class MyClass
{
public $var1;
public $var2;
public $var3;
function __construct($params)
{
if(isset($params['var1']))
$this->var1 = $params['var1'];
if(isset($params['var2']))
$this->var3 = $params['var2'];
if(isset($params['var3']))
$this->var3 = $params['var3'];
}
}
but as i said theese are optional parameters. Some object’s will use it, some wont use it. My question is i’m defining it everytime even it’s used or not. Should i define it when it’s passed something like:
if(isset($params['var1']))
public $var1 = $params['var1'];
i’m quite newbie with php, just need to know what i’m doing at top is right?
edit:typo.
Answer by Sarfraz
You can create an array instead and store only those values that are submitted like this:
class MyClass
{
public $data = array();
function __construct($params)
{
if(isset($params['var1']))
$this->data['var1'] = $params['var1'];
if(isset($params['var2']))
$this->data['var2'] = $params['var2'];
if(isset($params['var3']))
$this->data['var3'] = $params['var3'];
}
}
This way you have a single concrete known variable you can refer to anytime to get needed data back.
Answer by Starx
I dont like the whole concept. Since you are defining a public variable, what is the entire point. Unless you are blocking this, we can easily initialize a public variable outside the class, like:
$obj = new classname();
$obj -> unknowpublicname = "somevaue";
Whatever you do, unless you restrict some rules, or change the access modifiers, it does not make any difference.
