PHP函数无法正常工作（回显字符串，简单）

I created a function to allow me to debug PHP scripts so long as a variable ($debug) is set to 1:

function debug($msg) {
    if ($debug == 1) {
        echo $msg;
    } else {
        return false;
    }
}

That way, at the top of my script (before the functions.php file is called), I write:

$debug = 1;

to enable debugging, then:

debug("Function executed: " . $data);

at certain points so that I know string values/whatever at that point, with the desired response being the message displayed upon the screen.

However, regardless of what the value of the $debug string is, I never see any echo'd statements.

How can this be achieved?

It's difficult to say because you provided too few data.

The reason can be that your $debug variable is not known inside a function. Because using globals is not adviced, consider using constants define("DEBUG",1);.

EDIT

I presented within another question how I use a class for doing the same thing as class names are also globally accessed.

Debug is not available to your function because it is out of scope. You either:

1. Need to pass it as a parameter
2. Use the global keyword to include it in your function (discouraged)

.

function debug($msg, $debug){
    if($debug==1){
    echo $msg;
} else {
    return false;
    }
}

or

function debug($msg){
    global debug;
    if($debug==1){
    echo $msg;
} else {
    return false;
    }

}

The global variable is not accessible to functions until you make it so. Eg:

function debug($msg( {
   global $debug;
   etc...

PS: Please, don't do this. Find a better way. Really.

$debug is your global variable, so it is not accessible in your function

There is also the possibility to declare a const, (and then just insure the namespace is correct), like this:

const debug = true;
function newPrint($msg) {
  if (\debug === true) {
      echo $msg;
  } else {
      return false;
  }
}

ewPrint("heey");//will print "heey"

so just dont use a variable, but use a const.