在php类中包含文件

Question

我第一次在PHP中使用类做了一些事情.我想return object array在课堂上取一件物品.

这是我的班级

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;
            }
        }
    }
}

这是我的config.php档案

return (object) array(
    'host' => 'localhost',
    'username' => 'my_user',
    'pass' => 'my_pass',
    'database' => 'my_db'
);

我该怎么做？

PHP Parse error:  syntax error, unexpected '$dbconn' (T_VARIABLE)

Answer 1

您不能在变量定义中使用可执行代码,只能使用静态值.所以不支持这种事情:

class foo {
    public $var = result_of_some_function();
}

如果要初始化值,请使用构造函数.您可能最好将其作为配置文件阅读:

class User {
    public function __construct() {
        $config = json_decode(file_get_contents('config.json'));
        $conn = new mysqli($config->host, ...);
    }
}

或者更好,使用依赖注入:

class User {
    protected $db = null;
    public function __construct($db) {
        $this->db = $db;
    }
}

然后在您的代码中创建一个用户对象:

$db = new Db($config);
$user = new User($db);