JavaPersistenceWithHibernate第二版笔记-第六章-Mapping inheritance-002Table per concrete class with implicit polymorphism(@MappedSuperclass、@AttributeOverride)

一、结构



二、代码

1.

package org.jpwh.model.inheritance.mappedsuperclass;

import javax.persistence.MappedSuperclass;
import javax.validation.constraints.NotNull;

@MappedSuperclass
public abstract class BillingDetails {

@NotNull
protected String owner;

// ...

protected BillingDetails() {
}

protected BillingDetails(String owner) {
this.owner = owner;
}

public String getOwner() {
return owner;
}

public void setOwner(String owner) {
this.owner = owner;
}
}

2.

package org.jpwh.model.inheritance.mappedsuperclass;

import org.jpwh.model.Constants;

import javax.persistence.AttributeOverride;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.validation.constraints.NotNull;

@Entity
@AttributeOverride(
name = "owner",
column = @Column(name = "CC_OWNER", nullable = false))
public class CreditCard extends BillingDetails {

@Id
@GeneratedValue(generator = Constants.ID_GENERATOR)
protected Long id;

@NotNull
protected String cardNumber;

@NotNull
protected String expMonth;

@NotNull
protected String expYear;

// ...

public CreditCard() {
super();
}

public CreditCard(String owner, String cardNumber, String expMonth, String expYear) {
super(owner);
this.cardNumber = cardNumber;
this.expMonth = expMonth;
this.expYear = expYear;
}

public Long getId() {
return id;
}

public String getCardNumber() {
return cardNumber;
}

public void setCardNumber(String cardNumber) {
this.cardNumber = cardNumber;
}

public String getExpMonth() {
return expMonth;
}

public void setExpMonth(String expMonth) {
this.expMonth = expMonth;
}

public String getExpYear() {
return expYear;
}

public void setExpYear(String expYear) {
this.expYear = expYear;
}

}

3.

package org.jpwh.model.inheritance.mappedsuperclass;

import org.jpwh.model.Constants;

import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.validation.constraints.NotNull;

@Entity
public class BankAccount extends BillingDetails {

@Id
@GeneratedValue(generator = Constants.ID_GENERATOR)
protected Long id;

@NotNull
protected String account;

@NotNull
protected String bankname;

@NotNull
protected String swift;

public BankAccount() {
super();
}

public BankAccount(String owner, String account, String bankname, String swift) {
super(owner);
this.account = account;
this.bankname = bankname;
this.swift = swift;
}

public Long getId() {
return id;
}

public String getAccount() {
return account;
}

public void setAccount(String account) {
this.account = account;
}

public String getBankname() {
return bankname;
}

public void setBankname(String bankname) {
this.bankname = bankname;
}

public String getSwift() {
return swift;
}

public void setSwift(String swift) {
this.swift = swift;
}
}

三、存在的问题

1.it doesn’t support polymorphic associations very well。You can’t have another entity mapped with a foreign key “referencing BILLINGDETAILS ”—there is no such table. This would be problematic in the domain model, because BillingDetails is associated with User ; both the CREDITCARD and BANKACCOUNT tables would need a foreign key reference to the USERS table. None of these issues can be easily resolved, so you should consider an alternative mapping strategy.

2.查父类时要查每个表。Hibernate must execute a query against the superclass as several SQL SELECT s, one for each concrete subclass. The JPA query select bd from BillingDetails bd requires two SQL statements:

select
ID, OWNER, ACCOUNT, BANKNAME, SWIFT
from
BANKACCOUNT
select
ID, CC_OWNER, CARDNUMBER, EXPMONTH, EXPYEAR
from
CREDITCARD

3.several different columns, of different tables, share exactly the same semantics.

结论：We recommend this approach (only) for the top level of your class hierarchy,where polymorphism isn’t usually required, and when modification of the superclass in the future is unlikely.