PHP function scope [duplicate]

T开发者_C百科his question already has answers here: Reference: What is variable scope, which variables are accessible from where and what are "undefined variable" errors? (3 answers) Closed 2 years ago.

I know this is kind of a hacky thing I'm trying here, but I'm wondering if this is even possible. is there anyway I can access the $x variable in the following code, without passing arguments?

function foo() {
 $x = 1;
 return bar();
}

function bar() {
  //get $x from foo here somehow?
  return $x ? 'stuff' : 'other stuff';
}

---

I am not sure why, but you can use globals that opens up to a whole thing, but i will show you:

function foo() {
 global $x;
 $x = 1;
 return bar();
}

function bar() {
  global $x;
  //get $x from foo here somehow?
  return $x ? 'stuff' : 'other stuff';
}

Here is the demo: http://codepad.org/fPqUXzyC

It is always better to not use globals and just pass parameters, but if you cannot you could use globals

---

class baz {
   private $x;
   public function foo() {
      $this->x = 1;
      return $this->bar();
   }

   public function bar() {
      return $this->x ? 'stuff' : 'other stuff';
   }
}

---

You could make foo() store the $x value to $GLOBALS[] or global $x;. Other than that, nothing I can think of would do it. It would need to be purposely exposed to get it from inside another function.

If this is your code, I may recommend thinking about taking an object-oriented approach.

class Foo
{
  public static $x;

  public static function Foo(){
    Foo::$x = 1;
    return Foo::Bar();
  }

  public static function Bar() {
    return Foo::$x ? 'stuff' : 'other stuff';
  }
}

echo Foo::Foo();

Alternatively, do as others have suggested and pass $x as a function parameter.

---

I know this is kind of old, and an answer has already been accepted, but wouldn't this work?

function foo() 
  {  
    $x = 1;  
    return bar($x); 
  }  

function bar($x) 
  {
    return $x ? 'stuff' : 'other stuff'; 
  }