在功能WiTHOUT之外使用函数变量调用函数[关闭]

Question

I have a need for this. Is it possible?

I tried the following but it doesn't work:

$test;

function func_name() {
    global $test;
    $test = 'string';
}

echo $test; // I get nothing

Answer 1

If you don't call the function, nothing will happen.

You need to add func_name(); before echo $test;