mysqli_connect file for reuse

Question

I am trying to create one mysqli_conncet file for reuse. Other files/ functions are not able to run the function. Thank you.

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() . ")"
            );
    }   
    }
?>

My other files where I am calling the function:

<?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

a) $connect = ... is useless because (void) connect2db()

b) $connection is global. You have to mention it in insert_new_account() like global $connection;

c) Stop using `global` in PHP