如何从JAX-RS获取原始请求的URI javax.ws.rs.core.Response

Question

我正在尝试为JAX-RS编写一个定制的ExceptionMapper类。我想以某种方式读取来自JAX-RS javax.ws.rs.core.Response对象的原始请求URI。

当我检查与在调试模式下的IntelliJ IDEA响应对象然后我可以看到下面的路径下此信息：响应>上下文> resolvedUri其中
类型context的是org.glassfish.jersy.client.ClientResponse。这个类有一个resolvedUri变量，它拥有我需要的信息。

我能以某种方式获得这些信息吗？我如何写我的getRequestUri(response)方法？

public class MyExceptionMapper implements ExceptionMapper<WebApplicationException> { 

    @Override 
    public Response toResponse(WebApplicationException error) { 

     Response response = error.getResponse(); 
     ErrorResponse errorResponse = ErrorResponseBuilder 
       .builder() 
       .httpStatus(getDefaultStatusCodeIfNull(response)) 
       .errorMessage(getCustomErrorMessage(response)) 
       .requestedUri(getRequestedUri(response)) <--------- HOW TO READ IT? 
       .build(); 

     return Response 
       .status(errorResponse.getHttpStatus()) 
       .type(ExtendedMediaType.APPLICATION_JSON) 
       .entity(errorResponse) 
       .build(); 
    } 
} 

Answer 1

使用

@Context 私人的HttpServletRequest的servletRequest;

并使用HttpServletRequest.getRequestURI()

public class MyExceptionMapper implements 
    ExceptionMapper<WebApplicationException> { 

    @Context 
    private HttpServletRequest servletRequest; 

    @Override 
    public Response toResponse(WebApplicationException error) { 

     Response response = error.getResponse(); 
     ErrorResponse errorResponse = ErrorResponseBuilder 
       .builder() 
       .httpStatus(getDefaultStatusCodeIfNull(response)) 
       .errorMessage(getCustomErrorMessage(response)) 
       .requestedUri(servletRequest.getRequestURI()) 
       .build(); 

     return Response 
       .status(errorResponse.getHttpStatus()) 
       .type(ExtendedMediaType.APPLICATION_JSON) 
       .entity(errorResponse) 
       .build(); 
    } 
}