不能从另一个文件中使用php变量[重复]

Question

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• Reference: What is variable scope, which variables are accessible from where and what are “undefined variable” errors? 3 answers

like my title says, whenever I try to use my variables from another php file, it doesn't work (Undefined variable). I did declare them in the file that I'm including. For example, I have this file called variables.php that have this in it:

<?php
$DEBUG = TRUE;
$mysqli = new mysqli("127.0.0.1", "root", "", "29185917-database");
$DEBUG_LOG_FILE = "../log";
?>

And then I have another file called debug.php that tries to use the variable 'DEBUG' but it cannot access it. Here is my debug.php file:

<?php
require_once 'variables.php';
function echo_debug(string $message)
{
    if($DEBUG) {
        echo $message;
    }
}
?>

Whenever I try to use my function echo_debug I get the error message : Undefined variable 'DEBUG'. Any help on this problem is appreciated :).

</div>

Answer 1

Functions have their own scope. The variable is accessible, just not from inside the function.

You could pass $DEBUG as a parameter

function echo_debug(string $message, bool $DEBUG)

Then you would call it as

echo_debug("comment that will help me debug in dev mode", $DEBUG);

---

Another option is to declare DEBUG as a constant,

define('DEBUG', true);
$mysqli = new mysqli("127.0.0.1", "root", "", "29185917-database");
$DEBUG_LOG_FILE = "../log";

Then, in your function you would check for that constant:

function echo_debug(string $message) {
    if(DEBUG) { ... }
}

---

You could also use the global keyword, right above your if(), try to add global $DEBUG;.

require_once 'variables.php';
function echo_debug(string $message)
{
    global $DEBUG;
    if($DEBUG) { ... }
}

But generally the other two solutions are better, global variables are sometimes frowned upon.