$argv in a class

Question

I'm using exec() to exec a file but the file is in a class, I read more about argv but it was confusing. I need to get it to work inside a class.

It says:

Please note that, $argv and $argc need to be declared global, while trying to access within a class method

_{on php.net}

Samekh

And what the goal? Arguments are for console execution. – Hermanhermann

Answer 1

That means argc/argv aren't superglobals - they're only visible at the top-level context of PHP scripts, so...

<?php

$x = $argv[1]; // works

class foo {
   function bar() {
       echo $argv[1]; // undefined
   }
   function baz() {
       global $argv;
       echo $argv[1]; // works
   }
}

Lucilla

This can be done by using $GLOBALS['argv']... as well. In case copying a variable to your current scope or trying to make them superglobal is undesirable. – Isaiah

Answer 2

i add argv as arguments to the class constructor

class MyClass
{
    protected $args;

    public function __construct($args) 
    {
        $this->args = $args;
    }

    public function foo()
    {
        $this->args[1]; //access args using $this->args
    }
}

$myClass = new MyClass($args);

Adverse