I need allow types that implement two interfaces (Foo
and Bar
), not one of them.
interface Foo {};
interface Bar {};
class Foz implements Foo {};
class Baz implements Bar {};
class Foobar implements Foo, Bar {};
$resolver = new OptionsResolver();
$resolver->setRequired('data');
$resolver->setAllowedTypes('data', ['Foo', 'Bar']);
Wrong! allows Foz
and Baz
instances too.
I need allow types subclass of Bar
, not Bar
instances.
class Bar {};
class Foobar extends Bar {};
class FoobarBaz extends Foobar {};
$resolver = new OptionsResolver();
$resolver->setRequired('data');
$resolver->setAllowedTypes('data', ['Bar']);
Wrong! allows Bar
instances too.
I can redesign my classes/interfaces but it's not a design problem. So, is it possible achieve it with this component?
See Define a form option allowed values depending on another option value in a FormType.
You should use a normalizer for this:
use Symfony\Component\Form\Exception\InvalidConfigurationException;
$resolver->setNormalizer('data', function(Options $options, $data) {
if (!$data instanceof Foo && !$data instanceof Bar) {
throw new InvalidConfigurationException('"data" option must implement "Foo" and "Bar" interfaces.');
}
return $data;
});