如何回应功能之外的东西？

In the following code I want to echo green outside of public function.

Public function lol(){
$green ="green";
}

for example i want to echo $green in the following code.

public function green(){
echo"this is $green";
}

You pass $green in as a parameter and echo the function return value:

function green($green) {
  return "This is ". $green;
}

echo green('green'); //Results in: This is green
echo green('yellow'); //Results in: This is yellow

try the following code :

function __construct() {
       parent::__construct();
       $green ="green";
   }

public function green(){
echo"this is $green";
}

Place the $green variable in constructor.

$green;

class s {

    public function lol() {
        $GLOBALS['green'] = "green";
    }

}

$instance = new s();
$instance->lol();
echo $green;

Use global variables with $GLOBALS:

function lol(){
    $GLOBALS['green'] ="green";
}
lol();
echo $green;