在PHP中使用类外的变量

Question

Assuming that I have 3 PHP files:

1. index.php:

   require('config.php');
   require('connect_db.php');
   new connect_db();
   

2. config.php:

   $config['db_host'] = 'localhost';
   $config['db_username'] = 'root';
   $config['db_password'] = '';
   $config['db_name'] = 'my_db';
   

3. connect_db.php:

   class connect_db{
       function __construct(){
          $this->conn = new mysqli($config['db_host'], $config['db_username'], $config['db_password'], $config['db_name']);
       }
   }
   

When I run above codes, I meet an error: "Undefined variable: config in....".
My question is: How can I use the $config variable inside "connect_db" class without including config.php file in connect_db.php file.

Thanks!

Answer 1

As mentioned in the comments, pass the $config variable into the constructor of the class.

new connect_db($config);

class connect_db{
    function __construct($config){
       $this->conn = new mysqli($config['db_host'], $config['db_username'], $config['db_password'], $config['db_name']);
    }
}

Hope it helps.