Why php variable scope not working in function

Question

I executed following code but php says:

Notice: Undefined variable: b in ..\..\..\demo.php on line 4 
Notice: Undefined variable: a in ..\..\..\demo.php on line 4

Php Code:

<?php
  $a='a';$b='b';
  function test(){
      echo $a.$b;
  }
  test(); // error
?>

But i changed the code to this:

<?php
  $a='a';$b='b';
  function test($a,$b){
      echo $a.$b;
  }
  test($a,$b); // ab
?>

Why $a and $b are undefined in first case, since i defined them before? Why parameters need to pass in php? It's not require in other like JavaScript.

Add global $a, $b; in your first function and it will work :) — HamZa
– HamZa, Commented Mar 1, 2013 at 19:34
PHP and Javascript doe not follow the same scoping rules. See php.net/manual/en/language.variables.scope.php — Deinumite
– Deinumite, Commented Mar 1, 2013 at 19:38

Adam Plocher · Accepted Answer · 2013-03-01 19:44:39Z

2

If variables are defined outside the function, you need to specify the global keyword. Such as:

<?php
$a='a';$b='b';
function test(){
    global $a, $b;
    echo $a.$b;
}
test(); // error
?>

But your second example is the recommended way of handling it, typically.

edited Mar 1, 2013 at 19:44

answered Mar 1, 2013 at 19:34

Adam Plocher

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Comments

Sturm · Accepted Answer · 2013-03-01 19:35:02Z

1

$a and $b in the first example you provided are attempting to access those variables respectively from the local scope not the global scope. You could try declaring them like this

 function test() {
     global $a, $b;
     echo $a . $b; //or $GLOBALS['a'].$GLOBALS['b'];
 }

and you will get the correct values.

answered Mar 1, 2013 at 19:35

Sturm

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Comments

user2112300 · Accepted Answer · 2013-03-01 19:51:10Z

0

Try this

$a = '101';
$func = function() use($a) {
    echo $a;
};

function func_2() {
    global $func;
    $a = 'not previouse a';
    $func();
}

func_2();

answered Mar 1, 2013 at 19:51

user2112300

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