Question

Hey so I got tired of writing

echo "varname message";
var_dump(variable);

So I wrote this

function debugger($var, $message) {
    echo $message;
    var_dump($var);
    echo "<br />";
}

Which seems to work fine, except when its in a function. Then its like it doesn’t know that there’s a function defined, because its defined outside of the function. Like so.

function blah() {
    $x = 2;
    debugger($x, "this is x");
}

Also, I don’t understand functions, I knew you cant reference something in a function outside of the function without returning it, but I didn’t know you couldn’t reference variables or functions outside of the function without setting them as parameters. I think I have this wrong though.

So one more thing, does that mean that the variables inside of a function don’t conflict with the ones outside a function unless its returned?

Answer 1

Your code should work. Something else is wrong in your code. Try being more precise about what is not working and try isolating your problem. As for your other questions:

Variables defined in a function are scoped to that function, and will not interfere with variables from other functions. They also cannot access variables defined outside the function. For example

$a = 5;
function foo() {
    echo $a; //This will not work
}

What you can do is use the global keyword to “include” variables into your function’s scope, if you don’t want to pass them as arguments:

$a = 5;
function foo() {
    global $a;
    echo $a;
}

Or since a certain version of PHP (not sure which, if anyone knows please edit), you can use use:

$a = 5;
function foo() use ($a) {
    echo $a;
}