PHP：公共職能變量是不可見的

Question

外這裏是示例代碼：

class PeoplePDF extends TFPDF { 
    // Create basic table 
    public $test; 
    public function CreateTable($header, $data) 
    { 

     foreach ($header as $col) { 
      $this->Cell($col[1], 8, $col[0], 1, 0, 'C', true); 
     } 
     $this->Ln(); 
     // Data 
     $this->SetFillColor(255); 
     $this->SetTextColor(0); 
     $this->SetFont('Arial','','9'); 

     $date = date('d/m/y', strtotime($row["SendDate"])); 

     //$allRowsCount = count($data); 

     $db = new PDO('mysql:host=localhost;dbname=Service;', 'root', '', array(PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES utf8")); 
     $sqlNum = $db->prepare("select * From actnumbers"); 
     $sqlNum->execute(); 
     $sqlNumResult = $sqlNum->fetchAll(PDO::FETCH_ASSOC); 

     $lastnumber = $sqlNumResult[0]["LastNumber"]; 
     //$this->test = $sqlNumResult[0]["ActNumber"]; 
     $this->test = "test"; 

     foreach ($data as $row) 
     { 
      $date = date('d/m/y', strtotime($row["SendDate"])); 
       $this->Cell($header[0][1], 6, $lastnumber, 1, 0, 'C', true);    
     } 


    } 
} 

$myObj = new PeoplePDF(); 
$test = $myObj->test; 
echo $test; //it's not visible. echoes nothing 

我知道我有一個非常愚蠢的錯誤。 我是這個新手。 ） 我想我沒有正確地聲明變量$ test; 我該怎麼做？ 在此先感謝

Answer 1

class PeoplePDF extends TFPDF { 
public $test; 

    public function CreateTable($header, $data){ 

    $this->test = "test"; 

    } 

} 

類外，你可以使用它像這樣

$myObj = new PeoplePDF(); 
$test = $myObj->test;