从对象谓词自动创建swaggerUrlParameters

如何解决从对象谓词自动创建swaggerUrlParameters

我有两个对象Animal和从Animal继承的Dog

class Animal
    String name;

class Dog extends Animal
    String otherName;

，我有一个通用的rest终结点，可以继承Animal的类。该端点具有版本3的详尽文档。

class RestApi <T extends Animal>
    @Operation(summary = "Get animals")
    @GetMapping(produces = "application/json")
    @ApiResponses(value = {
            @ApiResponse(responseCode = "200",description = "Success"),@ApiResponse(responseCode = "401",description = "Unauthorized"),@ApiResponse(responseCode = "403",description = "Forbidden"),@ApiResponse(responseCode = "404",description = "Not Found"),@ApiResponse(responseCode = "500",description = "Failure")})
    public SearchResult<T> getItems(@RestPageableHandler final Pageable pageable,@RestPredicateHandler Object predicate,final HttpServletRequest request) {
            return crudService.getItems((Predicate) predicate,pageable,request.getParameterMap());
    }

，在RestApi for Dog中，我只需要这样做：

public class ApiRestDog extands RestApi<Dog> {
}

但是如果我要显示Dog的getItems（）的详细文档参数，则需要重写此方法，并确切地添加包含该对象谓词的内容，该谓词最终是在编译过程中创建的querydsl对象。 覆盖的方法如下所示：

    @Override
    @Parameters({
            @Parameter(name = "name",description = "Filter dog by name",in = ParameterIn.QUERY),@Parameter(name = "otherName",description = "Filter dog by other name",})
    public SearchResult<Price> getItems(@RestPageableHandler final Pageable pageable,@RestPredicateHandler(root = Dog.class) Object predicate,@RequestParam(required = false) final String target,final HttpServletRequest request) {
        return super.getItems(pageable,predicate,request);
    }

问题是，我是否可以通过任何方式基于父类和谓词（无需在RestApiDog.class上重写getItems（））为swagger创建此@Parameters？