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hibernate - 如何在运行简单的 hibernate 应用程序时摆脱 'java.lang.IllegalArgumentException: Unknown entity'?

转载 作者:行者123 更新时间:2023-12-04 02:56:48 26 4
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我是 Hibernate 的新手。在使用它创建一个小应用程序时,我遇到了以下异常:

线程“main”中的异常 java.lang.IllegalArgumentException:未知实体:
model.Students at org.hibernate.ejb.AbstractEntityManagerImpl.persist(AbstractEntityManagerImpl.java:223)
在 Controller .Main.main(Main.java:50)

有人可以帮我吗?

实体类如下:

Other details:
NetBeans Version: 6.7.1
Hibernate : 3.2.5

实体学生
package model;  
import java.io.Serializable;
import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.OneToOne;

@Entity
public class Students implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;

private String name;

@OneToOne(cascade=CascadeType.ALL)
private Address address;

public Address getAddress() {
return address;
}

public void setAddress(Address address) {
this.address = address;
}

public String getName() {
return name;
}

public void setName(String name) {
this.name = name;
}


public Long getId() {
return id;
}

public void setId(Long id) {
this.id = id;
}
}

另一个实体类
package model;

import java.io.Serializable;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;
@Entity

public class Address implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;


private String city;


private String zip;

public String getCity() {
return city;
}

public void setCity(String city) {
this.city = city;
}

public String getZip() {
return zip;
}

public void setZip(String zip) {
this.zip = zip;
}

public Long getId() {
return id;
}

public void setId(Long id) {
this.id = id;
}
}

DAO 文件
package controller;  import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.EntityTransaction;
import javax.persistence.Persistence;
import model.Address;
import model.Students;
import org.hibernate.HibernateException;
public class Main {
public static void main(String arr[])
{
EntityManagerFactory emf = Persistence.createEntityManagerFactory("OneToOne2PU");
EntityManager em = emf.createEntityManager();
EntityTransaction tr= em.getTransaction();
try{

tr.begin();

Address add1 = new Address();
add1.setCity("pune");
add1.setZip("09");
Address add2 = new Address();
add2.setCity("mumbai");
add2.setZip("12");

Students s1 = new Students();
s1.setName("abc");
s1.setAddress(add1);
Students s2 = new Students();
s2.setName("xyz");
s2.setAddress(add2);

em.persist(s1);
em.persist(s2);

tr.commit();
emf.close();

}
catch(HibernateException e){
e.printStackTrace();

}
}

}

持久性.xml
<?xml version="1.0" encoding="UTF-8"?>    
<persistence version="1.0" xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd">
<persistence-unit name="OneToOnePU" transaction-type="JTA">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<jta-data-source>students</jta-data-source>
<exclude-unlisted-classes>false</exclude-unlisted-classes>
<properties>
<property name="hibernate.hbm2ddl.auto" value="create-tables"/>
<property name="hibernate.dialect" value="org.hibernate.dialect.DerbyDialect"/>
<property name="hibernate.connection.username" value="app"/>
<property name="hibernate.connection.password" value="app"/>
<property name="hibernate.connection.url" value="jdbc:derby://localhost:1527/StudentsData"/>
</properties>
</persistence-unit>
</persistence>

hibernate.cfg.xml 文件
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-configuration PUBLIC "-//Hibernate/Hibernate Configuration DTD 3.0//EN" "http://hibernate.sourceforge.net/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
<session-factory>
<property name="hibernate.dialect">org.hibernate.dialect.DerbyDialect</property>
<property name="hibernate.connection.driver_class">org.apache.derby.jdbc.ClientDriver</property>
<property name="hibernate.connection.url">jdbc:derby://localhost:1527/sample</property>
<property name="hibernate.connection.username">app</property>
<property name="hibernate.connection.password">app</property>
<mapping class="model.Students"/>
<mapping class="model.Address"/>
</session-factory>
</hibernate-configuration>

最佳答案

取决于项目结构,但可能通过在persistence-unit 元素下直接向persistence.xml 添加以下内容。

<class>model.Students</class>
<class>model.Address</class>

就像这样:
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="1.0" xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd">
<persistence-unit name="OneToOnePU" transaction-type="JTA">
<provider>org.hibernate.ejb.HibernatePersistence</provider>

<jta-data-source>students</jta-data-source>
<class>model.Students</class>
<class>model.Address</class>
<exclude-unlisted-classes>false</exclude-unlisted-classes>
<properties>
<property name="hibernate.hbm2ddl.auto" value="create-tables"/>
<property name="hibernate.dialect" value="org.hibernate.dialect.DerbyDialect"/>
<property name="hibernate.connection.username" value="app"/>
<property name="hibernate.connection.password" value="app"/>
<property name="hibernate.connection.url" value="jdbc:derby://localhost:1527/StudentsData"/>
</properties>
</persistence-unit>
</persistence>

顺便说一句,为什么要在persistence.xml 和hibernate.cfg.xml 中配置像hibernate.dialect 这样的属性?

关于hibernate - 如何在运行简单的 hibernate 应用程序时摆脱 'java.lang.IllegalArgumentException: Unknown entity'?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/7887481/

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