Question

是否可以将两个DAO组合成一个Service方法？

我想创建一个通用方法，该方法将根据输入参数选择正确的DAO。现在我想出的是一种从服务对象外部接受Dao的方法。但这需要在Controller中初始化适当的Dao，这有点难看...

测量只是PostgreSQL上具有单独表的Temperature.java和Humidity.java实体的接口。

@Service
public class MeasurementService {

@Autowired
private TemperatureDao temperatureDao;

@Autowired
private HumidityDao humidityDao;

public<T extends PagingAndSortingRepository<Measurement, Long>> void insertMeasurementForUser(String username, List<Measurement> measurements, T dao) {
        dao.saveAll(measurements);
}
}

TemperatureDao.java

@Repository
public interface TemperatureDao extends PagingAndSortingRepository<Temperature, Long> {

    @Query("select u from Temperature u where u.owner = ?1 order by u.id desc")
    List<Temperature> findLatestTemperatureForUser(User user, Pageable pageable);
}

HumidityDao.java

@Repository
public interface HumidityDao extends PagingAndSortingRepository<Humidity, Long> {
    @Query("select u from Humidity u where u.owner = ?1 order by u.id desc")
    List<Humidity> findLatestHumidityForUser(User user, Pageable pageable);
}

Temperature.java

@Entity
@Table(name = "temperature")
public class Temperature implements Measurement {

    @Id
    @Column(name = "id")
    @GeneratedValue(strategy = GenerationType.SEQUENCE)
    private Long id;

    @Column(name = "th1value")
    private Float th1Value;

    @Column(name = "timestamp")
    @NotNull
    private LocalDateTime timestamp;

    @ManyToOne
    @JoinColumn(name = "user_id")
    @NotNull
    private User owner;

    public Temperature() {
    }

    public Temperature(Float th1Value, LocalDateTime timestamp, User owner) {
        this.th1Value = th1Value;
        this.timestamp = timestamp;
        this.owner = owner;
    }

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }
    @JsonDeserialize(using = LocalDateTimeDeserializer.class)
    public LocalDateTime getTimestamp() {
        return timestamp;
    }

    @JsonSerialize(using = LocalDateTimeSerializer.class)
    public void setTimestamp(LocalDateTime timestamp) {
        this.timestamp = timestamp;
    }

    @Override
    public User getOwner() {
        return owner;
    }

    @Override
    public void setOwner(User owner) {
        this.owner = owner;
    }
}

Humidity.java

@Entity
@Table(name = "humidity")
public class Humidity {

    @Id
    @Column(name = "id")
    @GeneratedValue(strategy = GenerationType.SEQUENCE)
    private Long id;

    @Column(name = "hum1value")
    private Float hum1Value;

    @Column(name = "timestamp")
    @NotNull
    private LocalDateTime timestamp;

    @ManyToOne
    @JoinColumn(name = "user_id")
    @NotNull
    private User owner;

    public Humidity() {
    }

    public Humidity(Float hum1Value, LocalDateTime timestamp, User owner) {
        this.hum1Value = hum1Value;
        this.timestamp = timestamp;
        this.owner = owner;
    }

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    @JsonDeserialize(using = LocalDateTimeDeserializer.class)
    public LocalDateTime getTimestamp() {
        return timestamp;
    }

    @JsonSerialize(using = LocalDateTimeSerializer.class)
    public void setTimestamp(LocalDateTime timestamp) {
        this.timestamp = timestamp;
    }

    public User getOwner() {
        return owner;
    }

    public void setOwner(User owner) {
        this.owner = owner;
    }
}

有什么想法吗？

Answer 1

您可以编写一个Resolver模式以根据您的条件返回所需的dao。您的服务将使用解析器获取正确的路径。

public HellDao implements BaseDao {
    public void save();
}

public ByeDao implements BaseDao {
    public void save();
}

public DaoResolver {

   @Autowired
   private helloDao;

   @Autowired
   private byeDao;


  public BaseDao resolve(Object input) {
       //based on input return the correct dao
       BaseDao resolvedDao = null;
       switch(input.enum) {
          case Hello:
            resolvedDao = helloDao;
            break;
          case Hello:
            resolvedDao = byeDao;
            break;
          default:
            //decide something for default
       }
   return resolvedDao;
  }
}

public class MyService {

   @Autowired
   private DaoResolver daoResolver;

   public Object doSomething() {
       BaseDao dao = daoResolver.resolve(someObject);
       //you will get HelloDao or ByeDao based on the input
       dao.save();
   }

}

Answer 2

您可以使用9_Whg_Nr_221检查测量类型，因此无需泛型即可进行测量。

fs.rename

如何在Spring Boot的Service层中以通用方式将两个DAO与一个方法组合在一起

2 个答案: