Including file inside a class php - php

Question

I am doing something using classes in php for very first time.
I am trying to fetch an return object array items in class.
This is my class
class User {
$dbconn = include("config.php");
private $dbHost = $dbconn->host;
private $dbUsername = $dbconn->username;
private $dbPassword = $dbconn->pass;
private $dbName = $dbconn->database;
private $userTbl = 'users';
function __construct(){
if(!isset($this->db)){
// Connect to the database
$conn = new mysqli($this->dbHost, $this->dbUsername, $this->dbPassword, $this->dbName);
if($conn->connect_error){
die("Failed to connect with MySQL: " . $conn->connect_error);
}else{
$this->db = $conn;
}
}
}
}
This is my config.php file
return (object) array(
'host' => 'localhost',
'username' => 'my_user',
'pass' => 'my_pass',
'database' => 'my_db'
);
How do i do it?
PHP Parse error: syntax error, unexpected '$dbconn' (T_VARIABLE)

Answer 1

You can't have executable code in a variable definition, only static values. So this sort of thing is not supported:
class foo {
public $var = result_of_some_function();
}
If you want to initialize a value, use the constructor. You're probably better off reading it as a config file:
class User {
public function __construct() {
$config = json_decode(file_get_contents('config.json'));
$conn = new mysqli($config->host, ...);
}
}
Or better, using dependency injection:
class User {
protected $db = null;
public function __construct($db) {
$this->db = $db;
}
}
Then in your code that creates a user object:
$db = new Db($config);
$user = new User($db);

Answer 2

The other way is to define constants in config file and use them in class.
in config.php file
define('HOST', 'localhost');
define('USERNAME', 'my_user');
define('PASS', 'my_pass');
define('DATABASE', 'my_db');
In class file
include("config.php")
class User {
private $dbHost = HOST;
private $dbUsername = USERNAME;
private $dbPassword = PASS;
private $dbName = DATABASE;
private $userTbl = 'users';
function __construct(){
if(!isset($this->db)){
// Connect to the database
$conn = new mysqli($this->dbHost, $this->dbUsername, $this->dbPassword, $this->dbName);
if($conn->connect_error){
die("Failed to connect with MySQL: " . $conn->connect_error);
}else{
$this->db = $conn;
}
}
}
}

Answer 3

Include this:
$dbconn = include("config.php");
in your construct function.

Answer 4

Maybe you should change in your code to
function __construct()
{
//included db file
include 'config.php';
if (!isset($this->db))
{
//code here
}

Answer 5

Try this way of using it.
<?php
class User
{
private $dbconn = null;
private $dbHost;
private $dbUsername;
private $dbPassword;
private $dbName;
private $userTbl = 'users';
function __construct()
{
include 'config.php'; //included file in constructor
if (!isset($this->db))
{
$this->dbHost= $this->dbconn->host;
$this->dbUsername= $this->dbconn->username;
$this->dbPassword= $this->dbconn->pass;
$this->dbName= $this->dbconn->database;
// Connect to the database
$conn = new mysqli($this->dbHost, $this->dbUsername, $this->dbPassword, $this->dbName);
if ($conn->connect_error)
{
die("Failed to connect with MySQL: " . $conn->connect_error);
} else
{
$this->db = $conn;
}
}
}
}
Config.php
<?php
$this->dbconn= (object) array(
'host' => 'localhost',
'username' => 'my_user',
'pass' => 'my_pass',
'database' => 'my_db'
);