java - 如何获取映射 URL 的 JSON 结果

Question

假设我们有以下 Controller

@Controller
@RequestMapping("/Index")
public class ControllerClass {

  @RequestMapping("/Result")
  @ResponseBody
  public List<Integer> result(){
     List<Integer> result = new ArrayList<Integer>();
     result.add(1);
     result.add(2);
     return result;
  }
 }

现在我想将 URL“/Index/Result”的 JSON 结果存储到字符串中。或者简单地，在应用注释后存储 Controller 的 JSON 结果。请注意，它不适用于为此目的考虑的测试和网络服务问题。有什么想法吗？提前致谢。

Answer 1

你可以注入(inject) Jackson 的 ObjectMapper进入 Controller 以手动将结果序列化为 JSON，然后再通过 ResponseEntity 返回.

@Configuration
public class Config {

     @Bean
     public ObjectMapper objectMapper() {
         // returning a plain ObjectMapper, 
         // you can change this to configure the ObjectMapper as requiered
         return new ObjectMapper();
     }
}


@Controller
@RequestMapping("/Index")
 public class ControllerClass {

   @Autowired
   private ObjectMapper objectMapper;

   @RequestMapping(value="/Result", 
                   method=RequestMethod.GET, 
                   produces="application/json")
   @ResponseBody
   public ResponseEntity<String> result(){
     List<Integer> result = new ArrayList<Integer>();
     result.add(1);
     result.add(2);
     String jsonResult = objectMapper.writer().writeValueAsString(result);
     // here you can store the json result before returning it;
     return new ResponseEntity<String>(jsonResult, HttpStatus.OK);
   }
}

编辑:

您还可以尝试定义 HandlerInterceptor捕获您感兴趣的请求的响应正文。

@Component
public class RestResponseInterceptor implements HandlerInterceptor {

      public void postHandle(HttpServletRequest request, HttpServletResponse response, Object handler, ModelAndView modelAndView) {
          // inspect response, etc...
     }
}