如何用JAX-RS捕捉WebApplicationException

我想抓住 WebApplicationException javax.ws.rs.ext.ExceptionMapper

@GET
@Path("/saySomething")
public List<String> saySomething() {
    String response = EchoRestClient.ping();
    List<String> list = new ArrayList<>();
    list.add(response);
    list.add("okay");
    return list;
}

(1)这是调用另一个rest api的客户端类:

public class EchoRestClient {
    private static Client client = ClientBuilder.newClient();

    public static String ping() {
        String serviceUrl = PropertyReader.getProperty(ServiceUrl.ECHO_SERVICE);

        Response response = client
                .target(serviceUrl)
                .path("saySomething")
                .request(ExtendedMediaType.APPLICATION_UTF8)
                .get();

        if (response.getStatus() == Response.Status.OK.getStatusCode()) {
            return response.getEntity(String.class);
        }

        throw new WebApplicationException(response);
    }
}

以及我的自定义异常处理程序,它不捕捉上述抛出的异常:

@Provider
public class WebservletExceptionMapper implements ExceptionMapper<Exception> {
    @Override
    public Response toResponse(Exception exception) {

        System.out.println("caught exception");
        Response response;

        if (exception instanceof WebApplicationException) {
            response = ((WebApplicationException) exception).getResponse();
        } else {
            response = Response....build();
        }

        return response;
    }
}

public static String ping() {
    // same code then before

    WebApplicationException e = new WebApplicationException(response);
    throw new RuntimeException("xxxxxx", e);
}

我上面的代码运行良好,当我调用 saySomething 我的web浏览器中的rest方法。

EchoService rest(包含被调用的 ping rest方法)HTTP 404在第一种情况下未被捕获。我需要抛出RuntimeException,因为未捕获WebApplicationException(第二种情况)。

这里怎么了?

如果我抛出此异常,则可以捕获: throw new WebApplicationException(response.getStatus())

但这一点不起作用: throw new WebApplicationException(response)