使用@MappedSuperclass中的@Id复合主键创建JPA实体
我有一个JPA实体的类层次结构,基类是定义了一个ID的MappedSuperclass。我试图在子类中使用复合键,但这似乎不起作用 我的代码如下所示使用@MappedSuperclass中的@Id复合主键创建JPA实体,jpa,eclipselink,Jpa,Eclipselink,我有一个JPA实体的类层次结构,基类是定义了一个ID的MappedSuperclass。我试图在子类中使用复合键,但这似乎不起作用 我的代码如下所示 @MappedSuperclass public class BaseEntity implements Serializable { @Id @GeneratedValue(strategy = GenerationType.IDENTITY) @Basic(optional = false) @Col
@MappedSuperclass
public class BaseEntity implements Serializable {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Basic(optional = false)
@Column(name = "id")
protected Long id;
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
}
@Entity
@EntityListeners(EntityBaseListener.class)
@Inheritance(strategy=InheritanceType.JOINED)
@Table(name = "catalog_entity")
public class BaseCatalogEntity extends BaseEntity {
@Column(name = "created_at", nullable = false)
@Temporal(TemporalType.TIMESTAMP)
private Date createdAt;
@Column(name = "updated_at", nullable = false)
@Temporal(TemporalType.TIMESTAMP)
private Date updatedAt;
public void setCreatedAt(Date date)
{
createdAt = date;
}
public void setUpdatedAt(Date date)
{
updatedAt = date;
}
public Date getCreatedAt() {
return createdAt;
}
public Date getUpdatedAt() {
return updatedAt;
}
}
@Entity
@Table(schema = "student_catalog")
@IdClass(value = StudentCatalog.StudentCatalogPK.class)
public class StudentCatalog extends BaseCatalogEntity {
@Id
@Column(name = "name", nullable = false, length = 100)
private String name;
@Id
@Column(name = "version", nullable = false)
private Integer version;
@Column(name = "description" , length = 255)
private String description;
@Column(name = "vendor" , length = 50)
private String vendor;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public Integer getVersion() {
return version;
}
public void setVersion(Integer version) {
this.version = version;
}
public String getDescription() {
return description;
}
public void setDescription(String description) {
this.description = description;
}
public String getVendor() {
return vendor;
}
public void setVendor(String vendor) {
this.vendor = vendor;
}
public static class StudentCatalogPK implements Serializable {
private Long id;
private String name;
private Integer version;
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public Integer getVersion() {
return version;
}
public void setVersion(Integer version) {
this.version = version;
}
@Override
public boolean equals(Object obj) {
boolean result = false;
if(obj != null && (obj instanceof StudentCatalogPK)) {
StudentCatalogPK other = (StudentCatalogPK)obj;
result = (Objects.equals(this.id, other.id) && Objects.equals(this.name, other.name) &&
Objects.equals(this.version, other.version));
}
return result;
}
@Override
public int hashCode() {
return (27780 + (this.id != null ? this.id.hashCode() : 0) +
(this.version != null ? this.version.hashCode() : 0) +
(this.name != null ? this.name.hashCode() : 0));
}
}
}
我得到以下例外情况:
异常说明:组合主键规范无效。主键类[com.example.jpa.StudentCatalog$StudentCatalogPK]和实体bean类[class com.example.jpa.StudentCatalog]中的主键字段或属性的名称必须对应,并且它们的类型必须相同。此外,请确保已为XML中的相应属性指定了ID元素和/或在实体类的相应字段或属性上指定了@ID
我正在使用Eclipselink 2.5.1。有没有一种方法可以在不更改BaseEntity和BaseCalogEntity类的情况下使其工作?在JPA中,在子类中重新定义id是不合法的。这将导致表映射和多态查询中的歧义
当业务密钥用于DB标识时,扩展超类中定义的密钥是一个常见问题。我建议仅使用代理密钥(like)作为DB标识,使用业务密钥作为实例标识。在以下条件下:
- 您的基本实体应该使用
表每类继承(正如我所看到的)
- 您的基本实体(复合键)键的类型与您希望在派生类中拥有的键的类型相同(因此还应该有String和Integer的复合键)
@AttributeOverride
注释,从中删除@Id
字段:
@AttributeOverride(name = "id", column = @Column(name = "NAME"))
因此,可以更改派生实体表中的列名,这是您所能达到的最大值。使用@MappedSuperClass时,建议将BaseEntity类设置为抽象类,然后从其他实体类扩展基类 更干净的方法—牢记继承并设计应用程序