如何回应功能之外的东西？

Question

在以下代码中，我想在echo之外green public function。

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

例如，我想在以下代码中回显$green。

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

Answer 1

尝试以下代码：

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

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

将$ green变量放在构造函数中。

Answer 2

您将$green作为参数传递并回显函数返回值：

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

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

Answer 3

$green;

class s {

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

}

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

Answer 4

将global variables与$GLOBALS：

一起使用

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