我的数据库包含3个表:用户和服务实体具有多对多关系,并与SERVICE_USER表连接,如下所示:
用户- SERVICE_USER -服务
SERVICE_USER表包含其他被阻止的列。
执行这种映射的最佳方式是什么?这些是我的实体类
@Entity
@Table(name = "USERS")
public class User implements java.io.Serializable {
private String userid;
private String email;
@Id
@Column(name = "USERID", unique = true, nullable = false,)
public String getUserid() {
return this.userid;
}
.... some get/set methods
}
@Entity
@Table(name = "SERVICES")
public class CmsService implements java.io.Serializable {
private String serviceCode;
@Id
@Column(name = "SERVICE_CODE", unique = true, nullable = false, length = 100)
public String getServiceCode() {
return this.serviceCode;
}
.... some additional fields and get/set methods
}
我遵循了这个例子http://giannigar.wordpress.com/2009/09/04/m ..。使用-jpa/下面是一些测试代码:
User user = new User();
user.setEmail("e2");
user.setUserid("ui2");
user.setPassword("p2");
CmsService service= new CmsService("cd2","name2");
List<UserService> userServiceList = new ArrayList<UserService>();
UserService userService = new UserService();
userService.setService(service);
userService.setUser(user);
userService.setBlocked(true);
service.getUserServices().add(userService);
userDAO.save(user);
问题是hibernate持久化User对象和UserService对象。CmsService对象未成功
我尝试使用急切获取-没有任何进展
使用上面提供的映射,是否有可能实现我期望的行为?
也许有一些更优雅的方式来映射多对多的连接表和额外的列?
发布于 2011-02-26 21:51:17
由于SERVICE_USER表不是纯join表,但具有额外的功能字段(阻塞),因此您必须将其映射为一个实体,并将用户和服务之间的多对多关联分解为两个OneToMany关联:一个用户具有多个UserServices,一个服务具有多个UserServices。
您还没有向我们展示最重要的部分:实体之间关系的映射和初始化(即您遇到问题的部分)。所以我会告诉你它应该是什么样子的。
如果您将关系设置为双向的,那么您应该拥有
class User {
@OneToMany(mappedBy = "user")
private Set<UserService> userServices = new HashSet<UserService>();
}
class UserService {
@ManyToOne
@JoinColumn(name = "user_id")
private User user;
@ManyToOne
@JoinColumn(name = "service_code")
private Service service;
@Column(name = "blocked")
private boolean blocked;
}
class Service {
@OneToMany(mappedBy = "service")
private Set<UserService> userServices = new HashSet<UserService>();
}
如果您没有在您的关系上放置任何级联,那么您必须持久化/保存所有实体。虽然只有关系的拥有方(这里是UserService方)必须初始化,但确保两端都一致也是一种好的做法。
User user = new User();
Service service = new Service();
UserService userService = new UserService();
user.addUserService(userService);
userService.setUser(user);
service.addUserService(userService);
userService.setService(service);
session.save(user);
session.save(service);
session.save(userService);
发布于 2013-06-26 15:49:07
我在xml文件配置中寻找一种方法来映射带有额外列的多对多关联表和hibernate。
假设with有两个表'a‘和'c’,它们与名为'extra‘的列有多对多关联。因为我没有找到任何完整的例子,这是我的代码。希望它能有所帮助:)。
首先是Java对象。
public class A implements Serializable{
protected int id;
// put some others fields if needed ...
private Set<AC> ac = new HashSet<AC>();
public A(int id) {
this.id = id;
}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public Set<AC> getAC() {
return ac;
}
public void setAC(Set<AC> ac) {
this.ac = ac;
}
/** {@inheritDoc} */
@Override
public int hashCode() {
final int prime = 97;
int result = 1;
result = prime * result + id;
return result;
}
/** {@inheritDoc} */
@Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (!(obj instanceof A))
return false;
final A other = (A) obj;
if (id != other.getId())
return false;
return true;
}
}
public class C implements Serializable{
protected int id;
// put some others fields if needed ...
public C(int id) {
this.id = id;
}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
/** {@inheritDoc} */
@Override
public int hashCode() {
final int prime = 98;
int result = 1;
result = prime * result + id;
return result;
}
/** {@inheritDoc} */
@Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (!(obj instanceof C))
return false;
final C other = (C) obj;
if (id != other.getId())
return false;
return true;
}
}
现在,我们必须创建关联表。第一步是创建一个表示复杂主键(a.id、c.id)的对象。
public class ACId implements Serializable{
private A a;
private C c;
public ACId() {
super();
}
public A getA() {
return a;
}
public void setA(A a) {
this.a = a;
}
public C getC() {
return c;
}
public void setC(C c) {
this.c = c;
}
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((a == null) ? 0 : a.hashCode());
result = prime * result
+ ((c == null) ? 0 : c.hashCode());
return result;
}
@Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
ACId other = (ACId) obj;
if (a == null) {
if (other.a != null)
return false;
} else if (!a.equals(other.a))
return false;
if (c == null) {
if (other.c != null)
return false;
} else if (!c.equals(other.c))
return false;
return true;
}
}
现在让我们创建关联对象本身。
public class AC implements java.io.Serializable{
private ACId id = new ACId();
private String extra;
public AC(){
}
public ACId getId() {
return id;
}
public void setId(ACId id) {
this.id = id;
}
public A getA(){
return getId().getA();
}
public C getC(){
return getId().getC();
}
public void setC(C C){
getId().setC(C);
}
public void setA(A A){
getId().setA(A);
}
public String getExtra() {
return extra;
}
public void setExtra(String extra) {
this.extra = extra;
}
public boolean equals(Object o) {
if (this == o)
return true;
if (o == null || getClass() != o.getClass())
return false;
AC that = (AC) o;
if (getId() != null ? !getId().equals(that.getId())
: that.getId() != null)
return false;
return true;
}
public int hashCode() {
return (getId() != null ? getId().hashCode() : 0);
}
}
现在,是时候用hibernate xml配置映射我们所有的类了。
A.hbm.xml和C.hxml.xml (相同)。
<class name="A" table="a">
<id name="id" column="id_a" unsaved-value="0">
<generator class="identity">
<param name="sequence">a_id_seq</param>
</generator>
</id>
<!-- here you should map all others table columns -->
<!-- <property name="otherprop" column="otherprop" type="string" access="field" /> -->
<set name="ac" table="a_c" lazy="true" access="field" fetch="select" cascade="all">
<key>
<column name="id_a" not-null="true" />
</key>
<one-to-many class="AC" />
</set>
</class>
<class name="C" table="c">
<id name="id" column="id_c" unsaved-value="0">
<generator class="identity">
<param name="sequence">c_id_seq</param>
</generator>
</id>
</class>
然后是关联映射文件a_c.hbm.xml。
<class name="AC" table="a_c">
<composite-id name="id" class="ACId">
<key-many-to-one name="a" class="A" column="id_a" />
<key-many-to-one name="c" class="C" column="id_c" />
</composite-id>
<property name="extra" type="string" column="extra" />
</class>
下面是要测试的代码示例。
A = ADao.get(1);
C = CDao.get(1);
if(A != null && C != null){
boolean exists = false;
// just check if it's updated or not
for(AC a : a.getAC()){
if(a.getC().equals(c)){
// update field
a.setExtra("extra updated");
exists = true;
break;
}
}
// add
if(!exists){
ACId idAC = new ACId();
idAC.setA(a);
idAC.setC(c);
AC AC = new AC();
AC.setId(idAC);
AC.setExtra("extra added");
a.getAC().add(AC);
}
ADao.save(A);
}
发布于 2017-07-11 14:37:59
如前所述,对于JPA,为了有机会拥有额外的列,您需要使用两个OneToMany关联,而不是单个ManyToMany关系。您还可以添加具有自动生成的值的列;这样,如果有用,它可以作为表的主键。
例如,额外类的实现代码应该如下所示:
@Entity
@Table(name = "USER_SERVICES")
public class UserService{
// example of auto-generated ID
@Id
@Column(name = "USER_SERVICES_ID", nullable = false)
@GeneratedValue(strategy = GenerationType.IDENTITY)
private long userServiceID;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "USER_ID")
private User user;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "SERVICE_ID")
private Service service;
// example of extra column
@Column(name="VISIBILITY")
private boolean visibility;
public long getUserServiceID() {
return userServiceID;
}
public User getUser() {
return user;
}
public void setUser(User user) {
this.user = user;
}
public Service getService() {
return service;
}
public void setService(Service service) {
this.service = service;
}
public boolean getVisibility() {
return visibility;
}
public void setVisibility(boolean visibility) {
this.visibility = visibility;
}
}
https://stackoverflow.com/questions/5127129
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