mysqli_connect return to private var

Question

someone could help me understand why __construct() is not working as expected? Or why $link is being Undefined?

I got error:

Notice: Undefined variable: link in D:\wamp\www\OI_WORK\forms\Projeto_Interface\ONGOING\php\class.php on line 29

This is line 29:

$consulta = mysqli_query($this->link, $query) or die("Error consulta" . mysqli_error($link));   

$link is defined here:

class DB{
  private $link;
  private $_DEFINICOES;

the __construct():

function __construct(){
    $this->_DEFINICOES = array(
          'database_ip' => "xxxx",
          'database_user' => "xxx",
          'database_senha' => "xxx",
          'database_name' => "xxxx");

    $this->link = mysqli_connect( $this->_DEFINICOES['database_ip'], $this->_DEFINICOES['database_user'], $this->_DEFINICOES['database_senha'], $this->_DEFINICOES['database_name']) or die("Error " . mysqli_error($this->link));          
    }

I call this class DB inside a private function in class controleUsuario and this generate the above error

  private function logar(){
     $banco = new DB();
     $retorno = $banco->consultarDB("
        SELECT `idusuario`
        FROM usuariosbeta 
        WHERE idusuario = {$_SESSION['login']}");
     print ("numero de linhas" . $retorno->num_rows);
  }

here is the function that calls controleUsuario->logar()

public function salvaPost($post){
     /* se o post nao estiver nulo ou com valor false atribui ao session */
     if( !(($post == false) or ($post == null)) ){
       foreach ($post as $key => $valor){
           /* segurança post */                
           $_SESSION[$key] = $valor;
        }
        $this->segurancaLogin($valor);  
        $this->logar();
     }
  }

regards,

Answer 1

found the error with a little help from a expert friend

$consulta = mysqli_query($this->link, $query) or die("Erro consulta" . mysqli_error($this->link));  

wrong

mysqli_error($link));

rigth

mysqli_error($this->link));