Question

我有一个酒店实体对象，需要对其进行实现分页。我尝试通过PagingAndSortingRepository接口实现它。我在实体中没有列。我的解决方案适用于“目的地代码”和“城市”等某些属性，但是当我尝试针对“酒店代码”和“酒店名称”获取结果时，我会在日志中看到计数查询

 ("Hibernate: select count(hotel0_.destinationcode) as col_0_0_ from Hotel hotel0_ where hotel0_."HOTELNAME"=?")

仅执行，而不是实际的选择查询，没有错误或异常。对于像findByCity这样的工作方法，我可以在日志中看到两个计数，然后是Select查询。

以上所有4个实体中的类型均为String。我的主要实体是酒店，它有一个称为HotelPk的嵌入式ID，我认为它就像具有复合密钥的标准实体一样。所以我的存储库中有效的方法是

Page<Hotel> findByIdDestinationcode(String destinationCode, Pageable pageRequest);

和

Page<Hotel> findByCity(String state, Pageable pageRequest);

不起作用的是

Page<Hotel> findByIdHotelcode(String hotelCode, Pageable pageRequest);
Page<Hotel> findByHotelname(String hotelName, Pageable pageRequest);

我的存储库的签名是

public interface HotelRepository extends JpaRepository<Hotel, HotelPK>, PagingAndSortingRepository<Hotel, HotelPK> {
//...... methods are defined here.
}

任何对此的帮助将不胜感激。

@Entity
@NamedQuery(name="Hotel.findAll", query="SELECT h FROM Hotel h")
public class Hotel implements Serializable {
private static final long serialVersionUID = 1L;

/*@EmbeddedId
private HotelPK id;*/
@Id
private String destinationcode;

private String hotelcode;

@Size(max = 20, message = "Column CITY cannot be more than 20 
characters.")
private String city;

//  @NotNull(message = "Column HOTELNAME cannot be null.")
@Size(max = 50, message = "Column HOTELNAME cannot be more than 50 
characters.")
private String hotelname;

/**
 * @return the destinationcode
 */
public String getDestinationcode() {
    return destinationcode;
}

/**
 * @param destinationcode the destinationcode to set
 */
public void setDestinationcode(String destinationcode) {
    this.destinationcode = destinationcode;
}

/**
 * @return the hotelcode
 */
public String getHotelcode() {
    return hotelcode;
}

/**
 * @param hotelcode the hotelcode to set
 */
public void setHotelcode(String hotelcode) {
    this.hotelcode = hotelcode;
}

/**
 * @return the city
 */
public String getCity() {
    return city;
}

/**
 * @param city the city to set
 */
public void setCity(String city) {
    this.city = city;
}

/**
 * @return the hotelname
 */
public String getHotelname() {
    return hotelname;
}

/**
 * @param hotelname the hotelname to set
 */
public void setHotelname(String hotelname) {
    this.hotelname = hotelname;
}
}
@Service(value = "HotelDao")
public class HotelDaoImpl implements HotelDao {
@Resource
private HotelRepository hotelRepository;

@Override
public Page<Hotel> searchHotel(String hotelCode, String hotelName, String 
      destinationCode, Pageable pageRequest) throws Exception {
    if(hotelCode!=null){
        return hotelRepository.findByHotelcode(Tools.padRight(hotelCode, 
3), pageRequest);
    }

    if(destinationCode!=null){
        return 
hotelRepository.findByDestinationcode(Tools.padRight(destinationCode, 3), 
pageRequest);
    }

    return hotelRepository.findByHotelname(hotelName, pageRequest);
}
}

public interface HotelRepository extends JpaRepository<Hotel, String> {
Page<Hotel> findByHotelcode(String hotelCode, Pageable pageRequest);

Page<Hotel> findByDestinationcode(String destinationCode, Pageable 
pageRequest);
Page<Hotel> findByHotelname(String hotelname, Pageable pageRequest);
}

Answer 1

由于PagingAndSortingRepository<Hotel, HotelPK>已经对其进行了扩展，因此无需再次扩展JpaRepository<Hotel, HotelPK>。您可以检查其实现。

因此，这样定义您的存储库。

public interface HotelRepository extends JpaRepository<Hotel, HotelPK> {
//...... methods are defined here.
}

现在，要问您的问题，您可以执行以下操作。

  public Page<Hotel> findAll(Pageable pageable) {
    Hotel hotel = new Hotel();
    hotel.setHotelCode("HC");
    hotel.setHotelName("Hotel");
    return hotelRepository.findAll(Example.of(hotel),pageable);
  }

在您的服务层

PagingAndSortingRepository findByattributename问题

1 个答案: