Is it possible to access outer local variable in PHP?

You could probably use a Closure, to do just that...

Edit : took some time to remember the syntax, but here's what it would look like :

function foo(){    $l = "xyz";    $bar = function () use ($l)    {        var_dump($l);    };    $bar();}foo();

And, running the script, you'd get :

$ php temp.phpstring(3) "xyz"

A couple of note :

• You must put a ; after the function's declaration !
• You could use the variable by reference, with a & before it's name : use (& $l)

For more informations, as a reference, you can take a look at this page in the manual : Anonymous functions

You must use the use keyword.

$bar = function() use(&$l) {};$bar();

In the very very old PHP 5.2 and earlier this didn't work. The syntax you've got isn't a closure, but a definition of a global function.

function foo() { function bar() { } }

works the same as:

function foo() { include "file_with_function_bar.php"; }

If you execute function foo twice, PHP will complain that you've tried to re-define a (global) function bar.

You can read default value by:

function(){    return preg_match(                     "yourVar = \d+"                   ,  str_file_get_contents(functionFile)                  ,   arrayToPutFieldsValue           ); }

If You would use two functons in the same time - it's like someone's using a spoon and You want to take a food from that spoon - You'll waste a food or some of You will starv.Anyway - You would have to set a pointer somehow in a hard way.It's impossible to get any field from other function or class without calling it to life.Functions/methods are instance-like - they need to be called.

Share the common fields by accessing a global fields with synchronized functions.