java - Spring Boot Rest Controller:返回默认错误JSON

在我的API中，我想保护用户详细信息端点，以便普通登录用户只能访问其用户配置文件。为此，我正在编写控制器：

@RequestMapping(value = URL_USER + "/{id}", method = RequestMethod.GET)
@ResponseBody
public PersistentEntityResource get(PersistentEntityResourceAssembler persistentEntityResourceAssembler, @PathVariable Long id) {
    Authentication authentication = SecurityContextHolder.getContext().getAuthentication();

    ApplicationUser loggedInUser = applicationUserService.findByUsername(authentication.getName());
    ApplicationUser applicationUser = applicationUserService.findById(id);

    if (applicationUser.getId().equals(loggedInUser.getId())) {
        return persistentEntityResourceAssembler.toFullResource(applicationUser);
    }

    throw new IllegalAccessException();
}

我想返回默认的spring boot error json，而不是引发导致InternalServerExcetption的Exception，如下所示：

{
    "timestamp": "2019-05-08T11:42:23.064+0000",
    "status": 403,
    "error": "Forbidden",
    "message": "Access Denied",
    "path": "/user/2"
}

我希望有一个解决方案，该解决方案对其他类似404的Erros一样有效。实现该目标的最简单方法是什么？

最佳答案

您可以使用以下方法进行相同的操作

public class FooController{

    //...
    @ExceptionHandler({ CustomException1.class, CustomException2.class })
    public String handleException() {
        return "the intended body";
    }
}

另外，您可以使用@ControllerAdvice将此逻辑作为全局异常处理程序放入

@ControllerAdvice
public class RestResponseEntityExceptionHandler
        extends ResponseEntityExceptionHandler {

    @ExceptionHandler(value
            = { IllegalArgumentException.class, IllegalStateException.class })
    protected ResponseEntity<Object> handleConflict(
            RuntimeException ex, WebRequest request) {
        String bodyOfResponse = "This should be application specific";
        return handleExceptionInternal(ex, bodyOfResponse,
                new HttpHeaders(), HttpStatus.CONFLICT, request);
    }
}