Файл mysqli_connect для повторного использования

Question

Я пытаюсь создать один файл mysqli_conncet для повторного использования. Другие файлы/функции не могут запускать функцию. Спасибо.

db_connect.php

<?php 
    define("DB_SERVER", "localhost"); 
    define("DB_USER", "***"); 
    define("DB_PASS", ""); 
    define("DB_NAME", "**"); 

    function connect2db(){ 
    global $connection; 
    $connection = mysqli_connect(DB_SERVER, DB_USER, DB_PASS, DB_NAME); 
     if (mysqli_connect_errno()) { 
     die("Database Connection failed!: " . 
      mysqli_connect_error() . 
      "(" .mysqli_connect_errno() . ")" 
      ); 
    } 
    } 
?> 

Мои другие файлы, в которых я вызываю функцию:

<?php 
    include_once ("../includes/db_connection.php"); 

    function insert_new_account(){ 

     $account_name = $_POST['acct_name']; 
     $account_address1 = $_POST['acct_address1']; 
     $account_address2 = $_POST['acct_address2']; 
     $account_city = $_POST['acct_city']; 
     $account_state = $_POST['acct_state']; 
     $account_zip = $_POST['acct_zip']; 
     $account_phone = $_POST['acct_phone']; 
     //$account_fax = $_POST['account_fax']; 

     $connect = connect2db(); 
     $query = mysqli_query($connection, "INSERT INTO accounts (account_name, account_address1, account_address2, account_city, account_state, account_zip, account_phone) 
     VALUES ($account_name, $account_address1, $account_address2, $account_city, $account_state, $account_zip, $account_phone)"); 

     return "Account {$account_name} has been created successfully."; 
    } 


    if (isset($_POST['create-new-account'])) { 
     insert_new_account(); 
    } 

?> 

Answer 1

а) $connect = ... бесполезно, потому что (void) connect2db()

б) $connection глобальна. Вы должны упомянуть его в insert_new_account() как global $connection;

с) Stop using `global` in PHP