运行此代码后的结果如下,有人能解释为什么名称没有正确传递,以至于它会说"Icefoots is years old"吗我是不是遗漏了什么?
年
<?php
class Penguin {
public function __construct($name) {
$this->species = 'Penguin';
$this->name = $name;
}
public function __toString() {
return $this->name . " (" . $this->species . ")'n";
}
public function getPenguinFromDb($id) {
// elegant and robust database code goes here
}
public function __get($field) {
if($field == 'name') {
return $this->username;
}
}
public function __set($field, $value) {
if($field == 'name') {
$this->username = $value;
}
}
public function __call($method, $args) {
echo "unknown method " . $method;
return false;
}
}
$tux = new Penguin('Icyfeet');
echo $tux->created;
echo $tux->name . " is " . $tux->age . " years old'n";
?>
我认为您试图访问用户名而不是名称。
public function __get($field) {
if($field == 'name') {
return $this->name;
}
}
好吧,在此之前,请声明类的所有字段,如:
private$名称=";private$species="企鹅";