正如一些用户建议的），返回mysqli实例

  function getConnected（$ host，$ user，$ pass，$ db）{

 $ mysqli = new mysqli（$ host，$ user，$ pass，$ db）; 

 if（$ mysqli-> connect_error）
 die（'Connect Error（'。mysqli_connect_errno（）。'）'。mysqli_connect_error（））; 

 return $ mysqli; 
} 
 

示例：

  $ mysqli = getConnected（'localhost'，'user'，'password'，'database'）; 
 

I've just started to look into Mysqli and I've understood most of it now, but I'm having a problem with a function that connects me to the database on other pages.

What I'm aiming to have is to just type getConnected(); where i need to connect.

this is the code:

function getConnected() {
$host = 'localhost';
$user = 'logintest';
$pass = 'logintest';
$db = 'vibo';

$mysqli = new mysqli($host, $user, $pass, $db);

if($mysqli->connect_error) {
    die('Connect Error (' . mysqli_connect_errno() . ') '
            . mysqli_connect_error());
}
}

and this is the error i get:

Notice: Undefined variable: mysqli in C:\xampp\htdocs\xampp\loginsystem\index.php on line 19

As some users have suggested (and is the best way), return the mysqli instance

function getConnected($host,$user,$pass,$db) {

   $mysqli = new mysqli($host, $user, $pass, $db);

   if($mysqli->connect_error) 
     die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());

   return $mysqli;
}

Example:

$mysqli = getConnected('localhost','user','password','database');