这个问题与我提出的其他question类似,但略有不同。
我有这个:
class A{
private List<B> bs;
...
}
class B{
private Long id;
private C c;
...
}
class C{
private Long id;
private String name;
...
}
我想要这个:
class A{
// the map should have b.c.name as key
private Map<String,B> bs;
...
}
class B{
private Long id;
private C c;
private A a;
...
}
class C{
private Long id;
private String name;
...
}
我不知道我是否清楚自己想要做什么,但它就像将一对多关系映射到一个地图一样简单,名称为C作为地图的关键,b为值。
提前致谢, Neuquino
答案 0 :(得分:4)
简短的回答是你不能这样做。
但这里的解决方案可能适合您的问题:
在实体A中,仍然将实体B定义为List(或者甚至更好地设置为Set,以便不能同时包含相同的B)。
@OneToMany(mappedBy="a")
private Set<B> bs;
由于您不想公开普通列表,请省略as
的getter和setter。
然后你可以为你的地图定义一个getter来实时构建地图:
// a transient field to cache the map
private transient Map<String, B> bsMappedByCName;
public Map<String, B> getBsMappedByCName() {
if(bsMappedByCName == null) {
bsMappedByCName = new HashMap<String, B>();
for(B b : bs) {
mapB(b);
}
}
// return as unmodifiable map so that it is immutable for clients
return Collections.unmodifiableMap(bsMappedByCName);
}
private void mapB(B b) {
// we assume here that field c in class B and likely also field name in class C are not nullable. Further more both of this fields sould be immutable (i.e. have no setter).
if(bsMappedByCName.put(b.getC().getName(), b) != null) {
// multiple bs with same CName, this is an inconsistency you may handle
}
}
要解决的最后一个问题是我们如何向A添加新B或删除一个。使用将地图返回为不可修改的策略,我们必须在A类中提供一些添加和卸载方法:
public void addB(B b) {
bs.add(b);
mapB(b);
}
public void removeB(B b) {
bs.remove(b);
bsMappedByCName.remove(b.getC().getName());
}
另一个选择是将return Collections.unmodifiableMap(...)
替换为(受ObservaleCollection from apache启发):
return new Map<String, B>() {
// implement all methods that add or remove elements in map with something like this
public B put(String name, B b) {
// check consistency
if(!b.getC().getName().equals(name)) {
// this sould be handled as an error
}
B oldB = get(name);
mapB(b);
return oldB;
}
// implement all accessor methods like this
public B get(String name) {
return bsMappedByCName.get(name);
}
// and so on...
};
答案 1 :(得分:0)
我有一个解决你的赏金的方法。它真的适合我。
这是一个玩具示例,在@CollectionOfElements上使用次要弃用警告。 您可以通过将其替换为取代它的较新注释来克服它。
package com.wladimir.hibernate;
import java.util.HashMap;
import java.util.Map;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinTable;
import org.hibernate.annotations.CollectionOfElements;
@Entity
public class A {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
private String name;
@CollectionOfElements
@JoinTable
private Map<String, B> bs = new HashMap<String, B>();
public A() {
}
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public Map<String, B> getBs() {
return bs;
}
public void setBs(Map<String, B> bs) {
this.bs = bs;
}
public void setName(String name) {
this.name = name;
}
public String getName() {
return name;
}
@Override
public String toString() {
return "A [bs=" + bs + ", id=" + id + ", name=" + name + "]";
}
}
package com.wladimir.hibernate;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.ManyToOne;
@Entity
public class B {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
private String name;
@ManyToOne
private C c;
@ManyToOne
private A a;
public B() {
}
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public C getC() {
return c;
}
public void setC(C c) {
this.c = c;
}
public A getA() {
return a;
}
public void setA(A a) {
this.a = a;
}
public void setName(String name) {
this.name = name;
}
public String getName() {
return name;
}
@Override
public String toString() {
return "B [a=" + a.getName() + ", c=" + c.getName() + ", id=" + id
+ ", name=" + name + "]";
}
}
package com.wladimir.hibernate;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
@Entity
public class C {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
private String name;
public C() {
}
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;
}
@Override
public String toString() {
return "C [id=" + id + ", name=" + name + "]";
}
}
package examples.hibernate;
import org.apache.commons.logging.Log;
import org.apache.commons.logging.LogFactory;
import org.hibernate.HibernateException;
import org.hibernate.SessionFactory;
import org.hibernate.cfg.AnnotationConfiguration;
public class PersistenceUtil {
private static final SessionFactory sessionFactory;
static {
Log log = LogFactory.getLog(PersistenceUtil.class);
try {
sessionFactory = new AnnotationConfiguration().configure().buildSessionFactory();
} catch (HibernateException ex) {
// Make sure you log the exception, as it might be swallowed
log.error("Initial SessionFactory creation failed.", ex);
throw ex;
}
}
public static SessionFactory getSessionFactory() {
return sessionFactory;
}
}
package examples.hibernate;
import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.Transaction;
import org.hibernate.criterion.Order;
import com.wladimir.hibernate.A;
import com.wladimir.hibernate.B;
import com.wladimir.hibernate.C;
import java.util.List;
import examples.hibernate.domain.*;
public class Tester{
private SessionFactory factory;
public Tester(SessionFactory factory) {
this.factory = factory;
}
@SuppressWarnings("unchecked")
public void runABC(String operation) {
Session session = factory.getCurrentSession(); // obtain/start unit of
// work
Transaction tx = null;
try {
tx = session.beginTransaction(); // start transaction
if ("Create".equals(operation)) {
A a = new A();
a.setName("A " + System.nanoTime());
C c = new C();
c.setName("C " + System.nanoTime());
session.save(c);
B b = new B();
b.setName("B " + System.nanoTime());
b.setA(a);
b.setC(c);
a.getBs().put(b.getName(), b);
session.save(b);
B b2 = new B();
b2.setName("B " + System.nanoTime());
b.setA(a);
b.setC(c);
a.getBs().put(b.getName(), b);
session.save(a);
} else if ("Read".equals(operation)) {
System.out.println("Reading data set.");
List<A> as = (List<A>) session.createCriteria(A.class)
.addOrder(Order.asc("name")).list();
for (A a : as) {
System.out.println(a);
}
}
tx.commit(); // commit transaction & end unit of work
} catch (RuntimeException ex) {
if (tx != null)
tx.rollback(); // abort transaction
throw ex;
}
}
// main application loop
public static void main(String args[]) throws Exception {
Tester app = new Tester(PersistenceUtil.getSessionFactory());
String operation = null;
if (args.length == 1) {
operation = args[0];
}
app.runABC(operation);
}
}
答案 2 :(得分:0)
如果这三个类都是实体,您可以尝试将其更改为Map&lt; String,C&gt; cs,从B关系中在C中放置一个反向字段,并将B-A关系改为C-A:
class A {
// the map should have c.name as key
@OneToMany
private Map<String, C> cs;
...
// This method is for getting the B for a key, like it was before.
public B getB(String key) {
return cs.get(key).getB();
}
}
class B {
private Long id;
@OneToOne // Or perhaps @ManyToOne.
private C c;
...
}
class C {
private Long id;
private String name;
@OneToOne(mappedBy="c") // Or perhaps maybe @OneToMany
private B b;
...
@ManyToOne(mappedby="cs")
private A a;
}
我假设这三个类是实体。如果C是Embeddable,那么我不知道该怎么做。