Class a {
public function __construct($a){
$this->age = $a;
}
}
Class b extends a {
public function printInfo(){
echo 'age: ' . $this->age . "\n";
}
}
$var = new b('age');
$var->printInfo();
我明白这段代码的工作原理,但是有可能将参数传递给类和父类的构造函数吗?是否可以将参数传递给类和父类构造函数?
下面我试图导致错误
Class a {
public function __construct($a){
$this->age = $a;
}
}
Class b extends a {
public function __construct($name){
$this->name = $name;
}
public function printInfo(){
echo 'name: ' . $this->name . "\n";
echo 'age: ' . $this->age . "\n";
}
}
$var = new b('name', 'age');
$var->printInfo();
?>
'父:: __结构();'B类将调用构造函数的类。所以试试'公共函数__construct($ name,$ age){$ this-> name = $ name;父:: __结构($岁); }' – Waygood 2013-04-30 07:07:21