难以将参数传递给函数？

Question

I'd like to pass it's value as an argument into my function

function dostuff($input) {
global $input;

    if ($input == 5) {
        $output = "Success";
    } else {
        $output = "Failure";
    }
echo $output;
}

Why does running

dostuff(5);

Not echo the output variable ("Success") like it's supposed to?

Answer 1

You don't need to set global $input;, try this:

function dostuff($input) {
    if ($input == 5) {
        $output = "Success";
    } else {
        $output = "Failure";
    }
echo $output;
}

With this, dostuff(5) return "Success"

Explanation:

global is to used global variable into your function (http://php.net/manual/en/language.variables.scope.php), you don't need it because you passed $input in function parameters.