我使用Apache CXF开发了Rest Application(JAX RS)。
CXF配置:
<jaxrs:server id="InternalRS" address="/V1">
<jaxrs:serviceBeans>
<ref bean="InternalService" />
</jaxrs:serviceBeans>
<jaxrs:providers>
<ref bean="jsonProvider" />
</jaxrs:providers>
<jaxrs:inInterceptors>
<bean id="CustomInterceptor"
class="com.test.web.interceptor.CustomInterceptor">
</bean>
</jaxrs:inInterceptors>
</jaxrs:server>
入境拦截器:
public class CustomInterceptor extends AbstractPhaseInterceptor<Message> {
public CustomInterceptor() {
super(Phase.READ);
}
@Override
public void handleMessage(Message message) throws Fault {
HttpServletRequest request = (HttpServletRequest) message.get(AbstractHTTPDestination.HTTP_REQUEST);
Transaction transaction = new Transaction();
String id = request.getHeader("id");
transaction.setId(id);
message.getExchange().put("transaction", transaction);
}
}
有没有办法让我可以通过使用出站拦截器修改JSON响应,将我的应用程序抛出的业务异常转换为等效的HTTP状态代码。
答案 0 :(得分:2)
与您的情况一样,由于某些条件,业务服务会抛出自定义异常。没有任何特殊处理,CXF将返回500错误并丢失自定义异常。这使得客户端模棱两可,并且不知道问题的确切原因。为了使它对客户端有用,您可以实现CXF异常处理程序。
您可以尝试使用ExceptionMapper来实现此目的。您需要为自定义BusinessException(或任何其他)和
创建ExceptionHandlerpackage com.gs.package.service;
public class ExceptionHandler implements ExceptionMapper<BusinessException> {
public Response toResponse(BusinessException exception) {
//you can modify the response here
Response.Status status;
status = Response.Status.INTERNAL_SERVER_ERROR;
return Response.status(status).header("exception", exception.getMessage()).build();
}
}
并在您的提供商列表中启用
<jaxrs:server id="sampleREST" address="/rest">
<jaxrs:serviceBeans>
<ref bean="yourSerivceBean">
</jaxrs:serviceBeans>
<jaxrs:providers>
<bean class="com.gs.package.service.ExceptionHandler"/>
</jaxrs:providers>
</jaxrs:server>