java Ho在响应对象中设置状态代码400

Question

我使用的是来自javax的响应对象。ws。rs api，即使我可以设置响应之类的状态。状态（响应、状态、错误请求）。build（），返回的响应代码始终为200，响应主体包含statusInfo参数，该参数表示BAD_请求。 有没有办法根据服务器端输入的一些验证将响应代码设置为400

Answer 1

不只是设置响应状态，而是从服务层抛出错误，让它传播到控制器。 在控制器中，你可以有

@Controller
public class MyService {
    @RequestMapping(value = "your api url path", method = RequestMethod.GET, produces = "application/
    public Response jsonAccounts(HttpServletRequest request,
            HttpServletResponse response) throws Exception {
        return Response.status(Response.Status.BAD_REQUEST).entity("Bad request received").build();
    }

    @ExceptionHandler(MyBadException.class)
    @ResponseStatus(value = HttpStatus.BAD_REQUEST, reason = "Bad request Request Exception")
    public void unexpectedError(HttpServletRequest request, Exception 
        log.debug("return HttpStatus.BAD_REQUEST, unexpectedError due to Exception", e);
    }
}

Answer 2

这对我很有效，我测试过

 @GET
@Path("/")
@Produces(MediaType.APPLICATION_JSON)
public Response getTestResult() {
    logger.info("Requested Test Service");
    try {
        Object result = "";[enter image description here][1]
      return  Response.status(Response.Status.BAD_REQUEST)
              .entity(result).build();
    } catch (Exception e) {
        return Response.ok("error", e.getMessage()).build();
    }

}