[英]Hibernate many One To Many relation in one class
i have three entity and Main(User) enitiy is in relation with other two entity,how can i retrive list of three entities from database in one query using hibernate 我有三个实体,并且Main(User)实体与其他两个实体有关,如何使用休眠在一个查询中从数据库中检索三个实体的列表
package hib.test;
import java.util.HashSet;
import java.util.Set;
public class Country {
private Integer id;
private String country;
private Set<User> userList = new HashSet<User>();
public Country() {
super();
// TODO Auto-generated constructor stub
}
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
public String getCountry() {
return country;
}
public void setCountry(String country) {
this.country = country;
}
public Set<User> getUserList() {
return userList;
}
public void setUserList(Set<User> userList) {
this.userList = userList;
}
}
User.java User.java
package hib.test;
public class User {
private Integer id;
private UserType userType;
private Country country;
public User() {
super();
// TODO Auto-generated constructor stub
}
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
public UserType getUserType() {
return userType;
}
public void setUserType(UserType userType) {
this.userType = userType;
}
public Country getCountry() {
return country;
}
public void setCountry(Country country) {
this.country = country;
}
}
UserType.java UserType.java
package hib.test;
import java.util.HashSet;
import java.util.Set;
public class UserType {
private Integer id;
private String userType;
private Set<User> userList = new HashSet<User>();
public UserType() {
super();
// TODO Auto-generated constructor stub
}
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
public String getUserType() {
return userType;
}
public void setUserType(String userType) {
this.userType = userType;
}
public Set<User> getUserList() {
return userList;
}
public void setUserList(Set<User> userList) {
this.userList = userList;
}
}
country.hbm.xml country.hbm.xml
<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<!-- Generated Jun 6, 2012 1:12:01 PM by Hibernate Tools 3.4.0.CR1 -->
<hibernate-mapping>
<class name="hib.test.Country" table="COUNTRY">
<id name="id" type="java.lang.Integer">
<column name="ID" />
<generator class="assigned" />
</id>
<property name="country" type="java.lang.String">
<column name="COUNTRY" />
</property>
<set name="userList" table="USER" inverse="false" lazy="true">
<key>
<column name="ID" />
</key>
<one-to-many class="hib.test.User" />
</set>
</class>
</hibernate-mapping>
user.hbm.xml user.hbm.xml
<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<!-- Generated Jun 6, 2012 1:12:01 PM by Hibernate Tools 3.4.0.CR1 -->
<hibernate-mapping>
<class name="hib.test.User" table="USER">
<id name="id" type="java.lang.Integer">
<column name="ID" />
<generator class="assigned" />
</id>
<many-to-one name="userType" class="hib.test.UserType" fetch="join">
<column name="USERTYPE" />
</many-to-one>
<many-to-one name="country" class="hib.test.Country" fetch="join">
<column name="COUNTRY" />
</many-to-one>
</class>
</hibernate-mapping>
usertype.hbm.xml usertype.hbm.xml
<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<!-- Generated Jun 6, 2012 1:12:01 PM by Hibernate Tools 3.4.0.CR1 -->
<hibernate-mapping>
<class name="hib.test.UserType" table="USERTYPE">
<id name="id" type="java.lang.Integer">
<column name="ID" />
<generator class="assigned" />
</id>
<property name="userType" type="java.lang.String">
<column name="USERTYPE" />
</property>
<set name="userList" table="USER" inverse="false" lazy="true">
<key>
<column name="ID" />
</key>
<one-to-many class="hib.test.User" />
</set>
</class>
</hibernate-mapping>
How can i retrive List<User>
, List<Country>
and List<UserType>
with one query 如何通过一个查询检索
List<User>
, List<Country>
和List<UserType>
EDIT 编辑
public static List<UserType> getUserTypeList() {
Session session = HibernateUtil.getSessionFactory().openSession();
Transaction transaction = null;
List<UserType> list = null;
try {
transaction = session.beginTransaction();
list = session.createQuery("from UserType as u").list();
if (list != null) {
for (UserType uType : list)
Hibernate.initialize(uType.getUserList());
}
transaction.commit();
} catch (Exception e) {
if (transaction != null)
transaction.rollback();
e.printStackTrace();
} finally {
session.close();
}
return list;
}
Actually I have 1 JTable and Two combo box each for UserType and Country So when i select any data in combo box JTable data should be filter according to selected value and in memory it shoud be save selected UserType and Country object. 实际上,我有1个JTable和2个组合框,分别用于UserType和Country。因此,当我在组合框中选择任何数据时,应根据所选值对JTable数据进行过滤,并且应在内存中保存所选的UserType和Country对象。
You need three queries to do that: 您需要执行以下三个查询:
select u from User u;
select ut from UserType ut;
select c from Country c;
EDIT: 编辑:
If what you actually want is a list of all the user types, with the users of each user type, and the country of each user of each user type, all this loaded in a single query, then you need fetch joins, as explained in the Hibernate documentation : 如果您真正想要的是所有用户类型的列表,以及每种用户类型的用户以及每种用户类型的每个用户的国家/地区,所有这些都加载在单个查询中,则您需要获取联接,如Hibernate文档 :
select userType from UserType userType
left join fetch userType.users user
left join fetch user.country
如果您想要一次关系数据并且内存不是问题,请执行lazy="false"
尝试在用户映射上将提取类型定义为EAGER,这样,在加载用户时,它将加载国家和用户类型。
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