What is the processing order of functions and strings?

Question

Summary

I want a function to execute only if a certain variable is true, however it checks the status of the variable before it exists.

At the top of my PHP page I have:

$debug = true;

At the bottom, among other functions I have:

function debug($message){
    echo ($debug ? "$message<br/>" : "");
}

And throughout the page I call:

debug("Debugging message here");

However, when I run the page with error reporting on, I get this error:

Notice: Undefined variable: debug in /examle.php on line 148

which references the middle line of the function.

How can I make a function that prints data only if the variable $debug, which is called at the top of the page, is true?

The variable `$debug` is [out of scope](http://www.php.net/manual/en/language.variables.scope.php) in the function `debug()`.... that's why we pass arguments to functions — Mark Baker, Jul 03 '15 at 08:16
You can either send `$debug` as a parameter or use `global $debug;` inside your function. I'd recommend not using `global` — Alex Tartan, Jul 03 '15 at 08:16
Why would you recommend not using `global`? Is there a third alternative that avoids me having to send it as a parameter each time the function is called? — Ben, Jul 03 '15 at 08:17

score 3 · Accepted Answer · answered Jul 03 '15 at 08:16

3

function debug($message){
    global $debug;
    echo ($debug ? "$message<br/>" : "");
}

You need to access the global var else it checks for a local one

answered Jul 03 '15 at 08:16

Sam

2

The "corecter" way would be to pass `$debug` into the function rather than making in global. Just saying. – Andrei Jul 03 '15 at 08:18
Why is making a variable global bad practice? – Ben Jul 03 '15 at 08:18
There's nothing wrong with making it global, obviously, the word `global` is there for a reason. It just hides dependencies, makes debugging a little harder too. But then again this is a little subjective. It's your call I guess. – Andrei Jul 03 '15 at 08:20