Question

在下面的代码示例中，我试图让类名Volvo出现在异常消息中，但出现的是CarException。在抛出异常时，如何在未指定__CLASS__的情况下将类名Volvo出现在异常消息中？

    class Volvo extends Car {
        public function smash() {
            throw new CarException("This car doesn't really smash.");
        }
    }

    abstract class Car {
        public function __construct() {
            var_dump(get_called_class());
        }
    }

    class CarException extends \Exception {
        public function __construct($message, $code = 0, Exception $previous = null) {
            $message = get_called_class() . ": $message";
            parent::__construct($message, $code, $previous);
        }
    }

    $volvo = new Volvo();
    try {
        $volvo->smash();
    } catch (CarException $e) {
        var_dump($e->getMessage());
    }

Answer 1

如果我理解你，你正在寻找getTrace：

class CarException extends \Exception {
    public function __construct($message, $code = 0, Exception $previous = null) {
        $trace = $this->getTrace();
        $cls = $trace[0]['class'];
        $message = $cls . ": $message";
        parent::__construct($message, $code, $previous);
    }
}

get_called_class不适用于此：它返回“拥有”代码的类的名称，而不是“调用”代码的类。

PHP扩展异常

1 个答案: