我在PHP方面还比较新手。
我想知道在另一个类的构造函数中实例化一个类并将其作为参数传递的正确方式。
(我在底部有一些测试代码)
class CryptoControl {
public $helper;
public $keyValues;
public function __construct ($primo1, $primo2, $message) {
$this->helper = new MathHelper();
$this->keyValues = new KeyValues($primo1, $primo2, $message);
}
public function criptografa_descriptografa_simetrica() {
$message = $keyValues->getMessage();
echo "Texto original: " . $message . "\n";
echo "Chave criptografada: " . base64_encode($message) . "\n";
$message_criptografada = base64_encode($message);
echo "Chave descriptografada: " . base64_decode($message_criptografada) . "\n";
}
}
$exec = new CryptoControl(17, 41, "TURING");
$resposta = $exec->criptografa_descriptografa_simetrica();
echo $resposta;
这就是我想要实例化的类:
class KeyValues {
public $numeroPrimo_1;
public $numeroPrimo_2;
public $message;
public function KeyValues ($primo1, $primo2, $message) {
$this->numeroPrimo_1 = $primo1;
$this->numeroPrimo_2 = $primo2;
$this->message = $message;
}
#getters and setters
}
我得到的错误是...
PHP 提示:在 /home/rodolfolottin/gitrepositories/Crypto/CryptoControl.php 的第72行中未定义变量:keyValues PHP 致命错误:在 /home/rodolfolottin/gitrepositories/Crypto/CryptoControl.php 的第72行上,对非对象调用了getMessage()函数
感谢您的帮助...