我有实体Person
如下:
@Entity
@Table(name = "Person")
@Inheritance(strategy = InheritanceType.JOINED)
public class Person implements Serializable {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@Column(name = "personName", length = 16, nullable = false)
private String name;
.....................
}
我有实体Teacher
如下:
@Entity
@Table(name = "teacher")
@PrimaryKeyJoinColumn(name = "PersonId")
public class Teacher extends Person implements Serializable {
private Map<String, Child> childByName = new HashMap<>();
......................................
}
我有实体Child
如下:
@Entity
@Table(name = "CHILD")
@PrimaryKeyJoinColumn(name = "PersonId")
public class Child extends Person implements Serializable {
..............................
}
map Teacher.childByName
应该映射name
Child
的属性(property)实体作为键和 Child
实体作为值。 Teacher
之间的关系和Child
是一对多。
我需要此映射采用 xml 格式。 现在我在 Person.hbm.xml 中执行此操作:
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-mapping PUBLIC
"-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
<class name="com.masterhibernate.SimpleHibernateDemo.Person"
table="Person">
<cache usage="read-write" />
<id name="id" column="id">
<generator class="native" />
</id>
<property name="name">
<column name="name" length="24" not-null="true" />
</property>
<property name="surname">
<column name="surname" length="36"></column>
</property>
<property name="address">
<column name="address" length="32"></column>
</property>
<joined-subclass name="com.masterhibernate.SimpleHibernateDemo.Parent"
table="Parent">
<key column="person_id" foreign-key="parent_person" />
<property name="job" column="WorkPlace" length="22" type="string" />
<set name="children" inverse="true" cascade="save-update" lazy="true">
<cache usage="read-write" />
<!-- specifies foreign key column of child table -->
<key column="ParentPK" />
<one-to-many class="com.masterhibernate.SimpleHibernateDemo.Child" />
</set>
</joined-subclass>
<joined-subclass name="com.masterhibernate.SimpleHibernateDemo.Child"
table="Child">
<key column="person_id" foreign-key="child_person" />
<property name="toy" column="toy" length="55" type="string" />
<many-to-one name="parent"
class="com.masterhibernate.SimpleHibernateDemo.Parent" column="ParentPK"
lazy="false" fetch="join" foreign-key="child_parent" />
<set name="teachers" table="TeacherPupil" inverse="true" lazy="true">
<key column="child_id" foreign-key="teacherPupil_child" />
<many-to-many class="com.masterhibernate.SimpleHibernateDemo.Teacher"
column="teacher_id" />
</set>
</joined-subclass>
<joined-subclass name="com.masterhibernate.SimpleHibernateDemo.Teacher"
table="Teacher">
<key column="person_id" foreign-key="teacher_person" />
<property name="subject" column="subject" length="25" type="string" />
<set name="children" table="TeacherPupil" lazy="false" fetch="join">
<cache usage="read-write" />
<key column="teacher_id" foreign-key="teacherPupil_teacher" />
<many-to-many class="com.masterhibernate.SimpleHibernateDemo.Child"
column="child_id" />
</set>
<map name="childByName" table="Child_By_Name" embed-xml="true">
<key column="childByName_id" />
<index column="childName" type="string" />
<one-to-many class="com.masterhibernate.SimpleHibernateDemo.Child" />
</map>
</joined-subclass>
</class>
</hibernate-mapping>
不幸的是<map>
甚至没有创建名为 Child_By_Name
的单独表。相反,它创建列 childByName
和childName
在表中Child
。这很奇怪。
那么,如何映射我的 childByName
map 要在单独的表格中引用Child
作为它的值?
最佳答案
我很感激所有的答案,但你还可以做的一件事是, 创建一个实体:
@Entity
@XmlRootElement(name = "ChildName")
@Table(name="ChildName")
public class ChildName{
@Id
private Long id;
private String name;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "childId")
private Child child;
}
并加入您的专栏
@Entity
@XmlRootElement(name = "Teacher")
@Table(name="Teacher")
public class Teacher {
@Id
private Long id;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "childNameId")
private ChildName childName;
}
你的整个代码会喜欢
@Entity
@XmlRootElement(name = "Teacher")
@Table(name="Teacher")
public class Teacher {
@Id
private Long id;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "childNameId")
private ChildName childName;
}
@Entity
@XmlRootElement(name = "ChildName")
@Table(name="ChildName")
class ChildName{
@Id
private Long id;
private String name;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "childId")
private Child child;
}
@Entity
@XmlRootElement(name = "Child")
@Table(name="Child")
public class Child extends Person implements Serializable {
}
@Entity
@XmlRootElement(name = "Person")
@Table(name="Person")
public class Person implements Serializable {
@Id
private Long id;
private String name;
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;
}
}
这不是 map 关联,它会像这样工作
关于java - 如何将 java.util.Map 与 hibernate xml 中的关联映射?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24217409/