我一遍又一遍地尝试在 MacOS 上的 NetBeans 11.2 中使用 JPA 时遇到以下异常。
java.lang.IllegalArgumentException：对象：javaapplication4.Id[id=5] 不是已知的实体类型。
在 org.eclipse.persistence.internal.sessions.UnitOfWorkImpl.registerNewObjectForPersist(UnitOfWorkImpl.java:4228)
在 org.eclipse.persistence.internal.jpa.EntityManagerImpl.persist(EntityManagerImpl.java:496)
在 javaapplication4.JavaApplication4.main(JavaApplication4.java:30)
我的主要课程：
package javaapplication4;
import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.EntityTransaction;
import javax.persistence.Persistence;
/**
*
* @author Informatica
*/
public class JavaApplication4 {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
EntityManagerFactory emf = Persistence.createEntityManagerFactory("JavaApplication4PU");
EntityManager em = emf.createEntityManager();
EntityTransaction tx = em.getTransaction();
tx.begin();
try{
Id id = new Id();
em.persist(id);
tx.commit();
}catch(Exception e ){
e.printStackTrace();
tx.rollback();
}
em.close();
emf.close();
}
}
我的实体类：
package javaapplication4;
import java.io.Serializable;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.NamedQueries;
import javax.persistence.NamedQuery;
import javax.persistence.Table;
/**
*
* @author Informatica
*/
@Entity
@Table(name = "ID")
@NamedQueries({
@NamedQuery(name = "Id.findAll", query = "SELECT i FROM Id i"),
@NamedQuery(name = "Id.findById", query = "SELECT i FROM Id i WHERE i.id = :id")})
public class Id implements Serializable {
private static final long serialVersionUID = 1L;
@javax.persistence.Id
@Column(name = "ID")
private Long id;
public Id() {
this.id = new Long(5);
}
public Id(Long id) {
this.id = id;
}
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
@Override
public int hashCode() {
int hash = 0;
hash += (id != null ? id.hashCode() : 0);
return hash;
}
@Override
public boolean equals(Object object) {
// TODO: Warning - this method won't work in the case the id fields are not set
if (!(object instanceof Id)) {
return false;
}
Id other = (Id) object;
if ((this.id == null && other.id != null) || (this.id != null && !this.id.equals(other.id))) {
return false;
}
return true;
}
@Override
public String toString() {
return "javaapplication4.Id[ id=" + id + " ]";
}
}
持久性.xml
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.1" xmlns="http://xmlns.jcp.org/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd">
<persistence-unit name="JavaApplication4PU" transaction-type="RESOURCE_LOCAL">
<provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
<class>javaapplication4.Id</class>
<properties>
<property name="javax.persistence.jdbc.url" value="jdbc:derby://localhost:1527/s"/>
<property name="javax.persistence.jdbc.user" value="s"/>
<property name="javax.persistence.jdbc.driver" value="org.apache.derby.jdbc.ClientDriver"/>
<property name="javax.persistence.jdbc.password" value="s"/>
<property name="javax.persistence.schema-generation.database.action" value="drop-and-create"/>
</properties>
</persistence-unit>
</persistence>
我的图书馆：
我正在使用 Derby DB。
让我知道您是否可以找到问题所在。先感谢您。