如何回应功能之外的东西？

Question

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";
}

Answer 1

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