Question

我正在尝试我的第一个OOP PHP，但我似乎遇到了障碍。据我所知，代码没有任何问题，我没有收到错误消息。谢谢！

<?PHP
class President {
    public $lastname;
    public $dob;
    public $dod;
    public $firstlady;
    public $wikipage;
    public $display;
    public $party;
    public $inoffice;

    public function __construct ($name, $dob, $dod, $girl, $wiki, $pic, $party, $terms){
        $this->lastname = $name;
        $this->dob = $dob;
        $this->dod = $dod;
        $this->firstlady = $girl;
        $this->wikipage = $wiki; 
        $this->display = $pic;
        $this->party = $party;
        $this->inoffice = $terms;
    }

    public function Write(){
        return "President". $name . "was in office for" . $terms . "." . "He was born on" . $dob . "and" . $dod . "." . "He represented the" . $party . "and lived in the Whitehouse with his wife" . 
        $girl . "<br/>" . "<img src'" . $pic . "' />" . "<br />" . "<a href'" . $wiki . "'> Read more on Wikipedia</a>";
    }
}

$obama = new President();
$obama->Write('Obama', 'June First 1992', 'is still alive', 'Michelle Obama', 'http://www.google.com', 'www.google.com/', 'Democrat', 'Two Terms');
echo $obama;

?>

Answer 1

首先，turn on错误报告...然后..

我可以看到2个问题。

首先，你构造函数需要很多参数，当你实例化对象时，你没有传递这些参数，但是当你调用Write方法时，它并不期望任何东西。

然后，当它返回一个字符串时，你不会回显$obama->Write(..)。

<强>溶液

$obama = new President('Obama', 'June First 1992', 'is still alive', 'Michelle Obama', 'http://www.google.com', 'www.google.com/', 'Democrat', 'Two Terms');
echo $obama->Write();

假设您的参数在构造中正确，

应该可以正常工作。

修改

如下面的评论中所述，您的write方法将无法访问任何类变量，因为它只查看本地范围。你需要改变你的变量调用：

return "President". $name

到

return "President". $this->name

OOP PHP不工作，网页没有返回任何＆gt;

1 个答案: