ApplicationContext::iShouldSeeValidJunitOutput PHP Метод

    public function iShouldSeeValidJunitOutput()
    {
        $dom = new \DOMDocument();
        $dom->loadXML($this->tester->getDisplay());
        if (!$dom->schemaValidate(__DIR__ . '/../..' . self::JUNIT_XSD_PATH)) {
            throw new \Exception(sprintf("Output was not valid JUnit XML"));
        }
    }