Question

我想获取发布到RestController的原始内容。我需要它对原始输入进行一些处理。

如何在不干扰过滤链的情况下获取原始内容？

Answer 1

以下是controllerAdvice的示例，您可以像在控制器中一样访问RequestBody和RequestHeader。 Model属性方法基本上是添加在所有页面或控制器流中使用的模型属性。它在控制器方法启动之前被调用。它提供了更清晰的方式来访问RESTful功能而不是复杂的方式。

@ControllerAdvice(annotations = RestController.class)
public class ControllerAdvisor {

    @ModelAttribute
    public void addAttributes(HttpServletRequest request, HttpServletResponse response,Model model, @RequestBody String requestString, @RequestHeader(value = "User-Agent") String userAgent) {
        // do whatever you want to do on the request body and header. 
        // with request object you can get the request method and request path etc.
        System.out.println("requestString" + requestString);
        System.out.println("userAgent" + userAgent);
        model.addAttribute("attr1", "value1");
        model.addAttribute("attr2", "value2");
    }   

}

Answer 2

我使用@ModelAttribute方法从@RequestBody设置值。

@ControllerAdvice 
public class CustomRestExceptionHandler extends ResponseEntityExceptionHandler 
{


    public CustomRestExceptionHandler() {
        super();
    }

    private Object request;

    @ModelAttribute
    public void setRequest(@RequestBody Object request) {
        this.request = request;
    }

    @Override protected ResponseEntity<Object> handleMethodArgumentNotValid(MethodArgumentNotValidException ex, HttpHeaders headers, HttpStatus status, WebRequest request) {
        logger.info(this.request)
    }

}

如何在Spring Boot Filter中获取HTTP Request主体内容？

2 个答案: