PHP OOP print a variable which has value inside the function but call from outside to be print

Question

Is it possible to print a variable which has the value inside the function but it's called from outside the function to be print in object oriented programming in PHP

Let's explain by example

My class looks like as:

class my {
   public $a;

   public function myFunc(){
       $name = "fahad";
       echo $this->a;
   }

}

It should print the value of $name when the function is call, as I am trying:

$class = new my();
$class->a = '$name';
$class->myFunc();

But it did't work and print the result as:

$name

I want it should print the value of variable $name which is inside the function

How it can be possible?

Thank You.

This is a bad idea. It violates abstraction. Why do you want to do this? — Will
– Will, Commented Jun 25, 2016 at 0:00
because i don't want to heavy my class functions and some data want to call from outside the function — Syed Aqeel
– Syed Aqeel, Commented Jun 25, 2016 at 0:32

Shira · Accepted Answer · 2016-06-25 00:13:34Z

2

You can use variable variables to do this, but it's usually considered bad practice.

class my {
   public $a;

   public function myFunc(){
       $name = "fahad";
       echo ${$this->a};
   }
}

$class = new my();
$class->a = 'name';
$class->myFunc();

Output:

fahad

answered Jun 25, 2016 at 0:13

Shira

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1 Comment

Syed Aqeel Over a year ago

Thank You very much this solved my biggest problem that is it what i wanted

Chin Leung · Accepted Answer · 2016-06-24 23:58:52Z

0

Inside your function, you can make a check:

public function myFunc(){
    if($this->a == '$name'){
        $name = 'fahad';
        echo $name;
    }else echo $this->a;
}

answered Jun 24, 2016 at 23:58

Chin Leung

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