我有麻烦映射一个类的嵌入式属性。我创建了一些类似于我试图说明的类。基本上,我有一个使用继承的@Embeddable类层次结构。顶级类“零件号”只有一个属性,扩展类不会为“零件号”类添加任何属性,它们只会添加一些验证/逻辑。JPA /休眠 - 嵌入属性
这里是我的意思是:
PART
@Entity
@Table(name="PART")
public class Part {
private Integer id;
private String name;
private PartNumber partNumber;
@Id
@GeneratedValue(strategy=GenerationType.SEQUENCE)
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
@Column(name="PART_NAME")
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
@Embedded
public PartNumber getPartNumber() {
return partNumber;
}
public void setPartNumber(PartNumber partNumber) {
this.partNumber = partNumber;
}
}
PARTNUMBER
@Embeddable
public abstract class PartNumber {
protected String partNumber;
private String generalPartNumber;
private String specificPartNumber;
private PartNumber() {
}
public PartNumber(String partNumber) {
this.partNumber = partNumber;
}
@Column(name = "PART_NUMBER")
public String getPartNumber() {
return partNumber;
}
public void setPartNumber(String partNumber) {
this.partNumber = partNumber;
}
/**
* @param partNumber
* @return
*/
public boolean validate(String partNumber) {
// do some validation
return true;
}
/**
* Returns the first half of the Part Number
*
* @return generalPartNumber
*/
@Transient
public String getGeneralPartNumber() {
return generalPartNumber;
}
/**
* Returns the last half of the Part Number
* which is specific to each Car Brand
*
* @return specificPartNumber
*/
@Transient
public String getSpecificPartNumber() {
return specificPartNumber;
}
}
FORD PARTNUMBER
public class FordPartNumber extends PartNumber {
/**
* Ford Part Number is formatted as 1234-#1234
*
* @param partNumber
*/
public FordPartNumber(String partNumber) {
super(partNumber);
validate(partNumber);
}
/*
* (non-Javadoc)
*
* @see com.test.PartNumber#validate(java.lang.String)
*/
@Override
public boolean validate(String partNumber) {
// do some validation
return true;
}
/*
* (non-Javadoc)
*
* @see com.test.PartNumber#getGeneralPartNumber()
*/
@Override
public String getGeneralPartNumber() {
return partNumber;
}
/*
* (non-Javadoc)
*
* @see com.test.PartNumber#getSpecificPartNumber()
*/
@Override
public String getSpecificPartNumber() {
return partNumber;
}
}
CHEVY PARTNUMBER
public class ChevyPartNumber extends PartNumber {
/**
* Chevy Part Number is formatted as 1234-$1234
*
* @param partNumber
*/
public ChevyPartNumber(String partNumber) {
super(partNumber);
validate(partNumber);
}
/*
* (non-Javadoc)
*
* @see com.test.PartNumber#validate(java.lang.String)
*/
@Override
public boolean validate(String partNumber) {
// do some validation
return true;
}
/*
* (non-Javadoc)
*
* @see com.test.PartNumber#getGeneralPartNumber()
*/
@Override
public String getGeneralPartNumber() {
return partNumber;
}
/*
* (non-Javadoc)
*
* @see com.test.PartNumber#getSpecificPartNumber()
*/
@Override
public String getSpecificPartNumber() {
return partNumber;
}
}
当然,这是不行的,因为Hibernate忽略了继承层次结构和不喜欢的事实,部分号码是抽象的。 有没有办法使用JPA或Hibernate Annotations来做到这一点?我曾尝试使用@Inheritance JPA注释。
我无法重构层次结构的“PartNumber”部分,因为原始开发人员希望能够使用N个XXXXPartNumber类扩展PartNumber。
有没有人知道这样的事情是否会成为JPA 2.0或新版本Hibernate的一部分?
尼斯通话。我很好奇如何发现子类,因为配置没有提及它们,并且JVM不允许你发现它们。 – skaffman 2009-07-13 19:20:21
“JVM不允许你发现它们”是什么意思? – Hardy 2009-07-14 08:39:56
@skaffman - 我想你需要将@Embedded标签移动到父类并使用@MappedSuperclass ...可能吗? – Petriborg 2009-07-14 19:19:12