JPA /休眠如何從一個實體加入特定字段到一個不同的實體

Question

假設我們已實體A和實體B，各有其相應的表中，A和B.

實體A是旅館，並且具有ID和幾個字段，例如國家，城市，郵政編碼等。

實體B是一個描述列表，它具有一個ID（與實體A中的ID相同）和另外兩個字段Language和Description。

我做下面的查詢在我的DAO：

Query query = getEntityManager().createQuery("FROM " + type.getSimpleName() + 
      " WHERE City = :cityParam AND Country = :countryParam"); 
    query.setParameter("cityParam", cityParam); 
    query.setParameter("countryParam", countryNameParam); 
    query.setMaxResults(numberOfResults); 
    List<HotelDto> hotelMap = query.getResultList(); 
    ListResponseModel result = new ListResponseModel(); 
    result.setHotelMap(hotelMap); 
    return result.getHotelMap().size() > 0 ? result : null; 

實體A，該酒店實體看起來是這樣的：

@Entity 
@Table(name = "propertyList", uniqueConstraints = 
{ @UniqueConstraint(columnNames = "HotelID")}) 
public class Hotel implements Serializable, EntityWithId<Long> { 
private static final long serialVersionUID = 1L; 

@Id 
@GeneratedValue 
@Column(name = "HotelID") 
private Long hotelId; 

@Column(name = "Name") 
private String name; 

@Column(name = "Address") 
private String address; 

@Column(name = "City") 
private String city; 

@Column(name = "StateProvince") 
private String stateProvince; 

@Column(name = "PostalCode") 
private String postalCode; 

@Column(name = "Country") 
private String country; 

@OneToOne(fetch = FetchType.LAZY) 
@PrimaryKeyJoinColumn(name = "propertyhotelid") 
private PropertyDescription propertyDescription; 
[...] 

實體B的描述，是這樣的：

@Entity 
@Table(name = "propertyDescriptionList", uniqueConstraints = 
{ @UniqueConstraint(columnNames = "HotelID")}) 
public class PropertyDescription implements Serializable, EntityWithId<Long> { 

private static final long serialVersionUID = 1L; 

@Id 
@Column(name = "HotelID", unique = true, nullable = false) 
private Long hotelId; 

@Column(name = "PropertyDescription") 
private String propertyDescription; 

問題是，一旦我得到結果集，一個實體將如下所示：

hotelId: 124125, 
name: "Random Hotel", 
address: "Some Address", 
city: "Shambala", 
postalCode: "W2 3NA", 
country: "Nevereverland", 
-propertyDescription: { 
    hotelId: 105496, 
    propertyDescription: "Insert Description here bla bla bla." 
} 
} 

我想是這樣的：

hotelId: 124125, 
name: "Random Hotel", 
address: "Some Address", 
city: "Shambala", 
postalCode: "W2 3NA", 
country: "Nevereverland", 
propertyDescription: "Insert Description here bla bla bla." 
} 

因爲我只是在說明本身它是一個字符串感興趣，而不是具有同樣的ID爲整個對象（一個副本）。

什麼是最好的方法來實現這一目標？

Answer 1

從你的問題聽起來像你正在尋找的是一種將兩張表映射到一個實體的方法？如果是這樣，你可以用@SecondaryTable來做到這一點：字符串'描述'然後將成爲你酒店實體的一個字段。

@Entity 
@Table(name = "propertyList", uniqueConstraints = { @UniqueConstraint(columnNames = "HotelID") }) 
@SecondaryTable(name = "propertyDescriptionList", pkJoinColumns = @PrimaryKeyJoinColumn(name = "HotelID"), uniqueConstraints = { @UniqueConstraint(columnNames = "HotelID") }) 
public class Hotel implements Serializable { 

    @Id 
    @Column(name = "HotelID") 
    private Long hotelId; 

    @Column(name = "PropertyDescription", table = "propertyDescriptionList") 
    private String propertyDescription; 

} 

Answer 2

您可以使用JPQL constructor expressions，並有這樣的事情

"select new my.package.HotelDTO(h) from Hotel h where ..." 

而且你會需要一個場String propertyDescription和HotelDTO

public HotelDTO(Hotel hotel) { 
    ... 
    this.propertyDescription = hotel.getPropertyDescription().getPropertyDescription(); 
    ... 
} 

Answer 3

構造爲什麼不把它作爲一個延遲訪場，而不是？

@Basic(fetch = FetchType.LAZY) 
private String propertyDescription; 

這樣，除非特別要求，否則不會提取它，而是將它作爲結果中的字符串字段獲取。