duanchou6534 2018-10-24 13:44
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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:

$mysqli = new mysqli("", "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:

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 :).

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3条回答 默认 最新

  • dqwcdqs358367 2018-10-24 13:49

    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("", "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.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?



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