php調用非對象的成員函數

Question

最後一行回顯不起作用。 我如何參考ID或暱稱？ 收到錯誤是調用一個成員功能的非物體上。 dsaojfj kjsaoj oisjao jpods JOP djopa jpaos JDO jpoajsoj運算PA POJ opdjaojs pojap jposajop jopajojdopsajodjop jaop jpoj ojdojpsoaj ojod jopajsdodjaospjop d添加細節.... SRR。

<?php 
class test{ 

    public $id; 
    public $nickname; 

    function __construct($id_, $nickname_) { 
     $this->id = $id_; 
     $this->nickname = $nickname_; 
    } 

    public function setId($id_){ 
     $this->id = $id_; 
    } 
    public function setNickname($nickname_){ 
     $this->nickname = $nickname_; 
    } 
    public function getId(){ 
     return $this->id; 
    } 
    public function getNickname(){ 
     return $this->nickname; 
    } 
} 

class InfoTest{ 

    public $tests = array(); 
    public $number; 

    function __construct() { 
     $this->number = 0; 
    } 

    public function addTest(test $test_){ 
     $this->number++; 
     $this->tests[$number] = $test_; 
    } 

    public function numberTests(){ 
     return $this->number; 
    } 

} 


$r = new test(1,2); 
$cc = new InfoTest; 
$cc->addTest($r); 
echo $cc->tests[1]->getId(); 

?> 

Answer 1

你第二次使用數量可變的忘了這 - $>：

$this->number++; 
    $this->tests[$number] = $test_; 

應該

$this->number++; 
    $this->tests[$this->number] = $test_;