如何访问其功能之外的变量

Question

if (isset($_GET['str'])) {
    fna();
} else { 
    fnb();
}

function fna(){
    // ...
    $result = 5;
}

function fnb(){
    // ... 
    $result = 9;
}

function fnc(){
    if ($result == 5){
        //error - undefined variable $result
    }
}

How can I get access to $result outside of fna() or fnb()?

Answer 1

This is how scope works in PHP: the variables named $result are all different instances, since they belong to the scopes of the functions in which they occur.

This is a good thing, and promotes good design.

You should return the value of your interest from fna and fnb instead of trying to use so-called side-effects. And then pass that value to your final function, so that it is clear it needs it. All this will make your code much more readable:

$result = isset($_GET['str']) ? fna() : fnb(); // <--- get return value

function fna(){
    return 5; // <--- return it
}

function fnb(){
    return 9;
}

function fnc($result){ // <--- what the function needs
    if ($result == 5){

    }
}

fnc($result); // <--- pass the value that the function needs.