Spring-通过异常处理程序拦截来自另一个bean的响应

Question

我有两个@一来控制器s-（A和B）并注册了响应EntityExceptionHandler。是否可以（以及如何做到）从A调用并在应用异常处理程序后从B获得响应？

例子：

1. 用户Rest呼叫A
2. A使用getPerson
调用
3. B引发异常未找到
4. NotFound由异常处理程序处理，转换ResponseEntity并放置400状态
5. B最终返回异常响应实体
6. A从B
获取400状态
7. 一个可以得到这个400并用它做点什么

简单的< code>@Autowired不起作用。

片段：

答:

@RestController
@RequestMapping("/v1")
public class A {

    private final B b;

    @Autowired
    public A(B b) {
        this.b = b;
    }

    @PostMapping(
        value = "persons",
        consumes = "application/json",
        produces = "application/json")
    public ResponseEntity<List<StatusResponse<Person>>> addPersons(final List<Person> persons) {
        final List<StatusResponse<Person>> multiResponse = new ArrayList<>();
        for(final Person p: persons) {
            final ResponseEntity<Person> response = b.addPerson(person);
            multiResponse.add(new StatusResponse<>(
                response.getStatusCode(), response.getMessage(), response.getBody()
            ));
        }
        return ResponseEntity.status(HttpStatus.MULTI_STATUS).body(multiResponse);
    }

}

B：

@RestController
@RequestMapping("/v1")
public class B {

    @PostMapping(
        value = "person",
        consumes = "application/json",
        produces = "application/json")
    public ResponseEntity<Person> addPerson(final Person person) {
        accessService.checkAccess();
        return ResponseEntity.status(201).body(
            logicService.addPerson(person)
        );
    }

}

处理程序

@ControllerAdvice
public final class MyExceptionHandler extends ResponseEntityExceptionHandler {

    @ExceptionHandler(MyException.class)
    protected ResponseEntity<Object> handleApiException(final MyException exception, final WebRequest webRequest) {
        //logic
        return afterLogic;
    }

}

Answer 1

转发类似于重定向，但完全发生在服务器端;Servlet 容器将相同的请求转发到目标 URL;网址不会在浏览器中更改。在Spring你可以这样做，你可以传递属性：

@Controller
@RequestMapping("/")
public class YourController {

    @GetMapping("/forward")
    public ModelAndView redirectWithUsingForwardPrefix(ModelMap model) {
        model.addAttribute("attribute", "forward");
        return new ModelAndView("forward:/redirectedUrl", model);
    }
}

Answer 2

不可能从异常处理程序将控制返回给控制器，异常处理程序是在处理其方法后调用的。您当前的流程看起来像这个调用A.add-

我不确定您想在控制器中处理此异常时做什么，但解决方案可能如下所示：

@RestController
@RequestMapping("/resources")
public class AController {

    private BService service;

    @Autowired
    public AController(BService service) {
        this.service = service;
    }

    @RequestMapping("/test")
    public ResponseEntity<String> test() {
        ResponseEntity<String> result = service.test();
        if (result.hasBody()) {
            //doSomething
        }

        return result; //or list like you did
    }
}

@Service
public class BService {

    public ResponseEntity<String> test() {
        try {
            return ResponseEntity.status(201).body(getResponse()); //this always throws exception. It's just for purpose of example
        } catch (CustomException ex) {
            return ResponseEntity.status(400).build();
        }

    }

    private String getResponse() {
        throw new CustomException("Not OK!");
    }
}