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');
$var->printInfo();
我明白這段代碼的工作原理,但是有可能將參數傳遞給類和父類的構造函數嗎?是否可以將參數傳遞給類和父類構造函數?
下面我試圖導致錯誤
Class a {
public function __construct($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');
$var->printInfo();
?>
'父:: __結構();'B類將調用構造函數的類。所以試試'公共函數__construct($ name,$ age){$ this-> name = $ name;父:: __結構($歲); }' – Waygood 2013-04-30 07:07:21