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我有许多具有相同方法的 Controller ,区别仅在于实体类。
我想创建通用的 BaseController 并且在使用 QuerydslPredicate 注释时遇到问题(无法设置根,这是通用的)
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);
}
}
方法 extractTypeInfo 从 QuerydslPredicateArgumentResolver 返回 T
。但需要实体类。
而且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?
最佳答案
我认为没有办法为注解提供通用类。它应该是编译时常量。
但是这个呢? (大量代码,完整示例)。
Spring 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;
}
Controller
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,您仍然需要为每个 @Entity
生成额外的 @Controller
和 @Repository
类。尽管它们大多是空 stub 。但我认为您无法完全欺骗 @Querydsl 或 Spring Data 以在联合中正确使用泛型。
关于java - QuerydslPredicate 和泛型类,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37481905/
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