我使用 JPA 和 Hibernate 作为提供程序,我想映射 Map<Embeddable, Double>
与 @ElementCollection ,但是当坚持时,我遇到了 hibernate 异常:
javax.persistence.PersistenceException: org.hibernate.InstantiationException: No default constructor for entity: java.lang.Double
我知道:
if the values of the Map are either embeddable or basic types, the Map is mapped as an element collection. [Pro JPA 2 Second edition- Page 102]
所以在我的例子中我应该使用@ElementCollection,但是为什么hibernate将Double视为一个实体并需要一个默认的构造函数?
Hibernate/JPA 依赖项:
<properties>
<hibernate.version>5.2.10.Final</hibernate.version>
</properties>
<dependencies>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-core</artifactId>
<version>${hibernate.version}</version>
</dependency>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-entitymanager</artifactId>
<version>${hibernate.version}</version>
</dependency>
<dependency>
<groupId>org.hibernate.javax.persistence</groupId>
<artifactId>hibernate-jpa-2.1-api</artifactId>
<version>1.0.0.Final</version>
</dependency>
....
<dependencies>
我的类(我使用带有继承单表策略的泛型类型):
@Entity
@Table(name = "GENERIC_STAFF")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name = "My_Discriminator", discriminatorType = DiscriminatorType.STRING, length = 255)
public class FirstLevelBaseClass {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
public Long getId() {
return id;
}
}
@Getter @Setter
public class SecondLevelBaseClass <K extends Serializable, V extends Serializable, M extends Map<K, V>> extends FirstLevelBaseClass{
private M value;
}
@Entity // Class that I want to persist
@DiscriminatorValue(value = "My_DoubleMapClass")
@Access(AccessType.PROPERTY)
public class DoubleMapClass extends SecondLevelBaseClass<EmbeddableDate, Double, SortedMap<EmbeddableDate, Double>>{
@Override
@ElementCollection
@CollectionTable(joinColumns = @JoinColumn(name = "ID_GENERIC"))
@AttributeOverrides({
@AttributeOverride(name = "key.start", column = @Column(name = "START")),
@AttributeOverride(name = "key.end", column = @Column(name = "END"))
})
@Column(nullable = false, name = "MyValue")
@OrderBy
public SortedMap<EmbeddableDate, Double> getValue() {
return super.getValue();
}
@Override
public void setValue(SortedMap<EmbeddableDate, Double> value) {
super.setValue(value);
}
}
最佳答案
如crizzis mention ,当我摆脱@AttributeOverride
时,我可以毫无异常(exception)地坚持下去(这似乎是 Hibernate 中的一个错误),因此为了保持覆盖键和值列名称的能力,我创建了 Double 的包装类为 @Embeddable ,所以我的 Collection 变成 SortedMap<EmbeddableDate, EmbeddableDouble>
.
另一个解决方案是不覆盖 EmbeddableDate
字段并直接在 EmbeddableDate
中命名它们的列,但这对我来说是不可能的,因为我正在使用 EmbeddableDate
在其他实体中(我将失去它的可重用性!)。
关于java - 为什么 hibernate 需要 java.lang.Double 的默认构造函数?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44393507/