Why does this simple code yield 'Undefined variable: bar'

Question

class A {
    protected $bar = 'bar';
    public function foo() {
        echo $this->$bar;
    }
}

$a = new A();
$a->foo();

It's boggling my mind that this isn't working. I come from C++ and C# so it's likely something I don't understand about PHP.

Change `echo $this->$bar;` to `echo $this->bar;` – asprin Nov 15 '13 at 14:23 — asprin, Nov 15 '13 at 14:23
http://php.net/manual/en/language.oop5.basic.php – kunal Nov 15 '13 at 14:30 — kunal, Nov 15 '13 at 14:30

Jason McCreary · Answer 1 · 2013-11-15T14:45:05.113

6

Why does this simple code yield 'Undefined variable: bar'

Because PHP tries to evaluate the variable $bar before evaluating $this->. Since there is not a $bar variable, it yields a notice.

Remove the $ in front of $bar:

echo $this->bar;

I would encourage you to read the Variable Variables section of the docs as well as OOP Basics.

edited Nov 15 '13 at 14:45

answered Nov 15 '13 at 14:23

Jason McCreary

I'm curious, why does dynamic language imply that specific order of evaluation? – Nick Strupat Nov 15 '13 at 14:41
1

It does not imply specific order of evaluation, more the variable variables. I mentioned it since you noted coming from C++. Which is far more static than PHP. Nonetheless, I have adjusted my answer. – Jason McCreary Nov 15 '13 at 14:44

score 2 · Accepted Answer · answered Nov 15 '13 at 14:23

2

When you access a member, you only need the dollar-sign before this; i.e. access it like this instead:

echo $this->bar;

answered Nov 15 '13 at 14:23

Peter Bloomfield

score 0 · Answer 3 · answered Nov 15 '13 at 14:30

0

try this

public function(){
  echo $this->bar;
}

When use the $this then couln't use the symbol $ before variable $this->$variable but $this->variable;

answered Nov 15 '13 at 14:30

A. Zalonis

3 Answers3