Querydsl谓词和泛型类

java spring-data generic-collections querydsl

2022-09-02 14:07:54

我有很多控制器使用相同的方法，区别只是实体类。

我想创建泛型 BaseController，但 Querydsl谓词 annotaion 有问题（无法设置根，这是泛型的）

class abstract BaseController<T extends BaseEntity> {
   @Autowired
   private Repository<T, Long> repository;

   @RequestMapping(method = RequestMethod.GET)
   public Page<T> findAll(@QuerydslPredicate Predicate predicate, Pageable pageable) {
      return repository.findAll(predicate, pageable);
   }
}

Method extractTypeInfo from QuerydslPredicateArgumentResolver return T.但需要实体类。

我无法将根值设置为QuerydslPredicate（没有类）

@QuerydslPredicate(root = T.class)

关于如何实现这一目标的任何帮助？

答案 1

我不认为有办法提供泛型类来注释。它应该是编译时常量。

但是这个怎么样呢？（大量代码，完整示例）。

弹簧底座

@SpringBootApplication
public class Application {

    public static void main(String[] args) {
        SpringApplication.run(Application.class, args);
    }
}

实体

@MappedSuperclass
@Data
@NoArgsConstructor
@AllArgsConstructor
public abstract class BaseEntity {

    @Id
    @Column(name = "id")
    private Integer id;

    @Column(name = "created_at")
    private Date createdAt;
}

@Entity
@Table(name = "user")
@Data
@NoArgsConstructor
@AllArgsConstructor
@EqualsAndHashCode(callSuper = true)
public class User extends BaseEntity {

    @Column(name = "username")
    private String userName;
}

@Entity
@Table(name = "another_user")
@Data
@NoArgsConstructor
@AllArgsConstructor
@EqualsAndHashCode(callSuper = true)
public class AnotherUser extends BaseEntity {

    @Column(name = "another_username")
    private String anotherUserName;
}

控制器

public abstract class BaseEntityController<T extends BaseEntity> {

    private final BaseEntityRepository<T> repository;

    @Autowired
    protected BaseEntityController(BaseEntityRepository<T> repository) {
        this.repository = repository;
    }

    @GetMapping(value = "/index")
    public Page<T> index(Predicate predicate, Pageable pageable) {
        return repository.findAll(predicate, pageable);
    }
}

@RestController
@RequestMapping("/user")
public class UserController extends BaseEntityController<User> {

    protected UserController(BaseEntityRepository<User> repository) {
        super(repository);
    }

    @Override
    public Page<User> index(@QuerydslPredicate(root = User.class) Predicate predicate, Pageable pageable) {
        return super.index(predicate, pageable);
    }
}

@RestController
@RequestMapping("/anotheruser")
public class AnotherUserController extends BaseEntityController<User> {

    protected AnotherUserController(BaseEntityRepository<User> repository) {
        super(repository);
    }

    @Override
    public Page<User> index(@QuerydslPredicate(root = AnotherUser.class) Predicate predicate, Pageable pageable) {
        return super.index(predicate, pageable);
    }
}

存储库

@NoRepositoryBean
public interface BaseEntityRepository<T extends BaseEntity> extends CrudRepository<T, Integer>, QuerydslPredicateExecutor<T> {
}

@Repository
public interface UserEntityRepository extends BaseEntityRepository<User> {
}

@Repository
public interface AnotherUserEntityRepository extends BaseEntityRepository<AnotherUser> {
}

它尽可能地工作和抽象。不过，你仍然需要生成额外的和职业。尽管它们大多是空的存根。但我不认为你可以完全欺骗@Querydsl或Spring Data在联合中正确使用泛型。@Controller@Repository@Entity

答案 2