Question

这是一个类似的帖子，我之前看到的关于这个例外，但我完全迷失了。我还没有使用JPA将实体持久化到数据库，尽管我已经从使用它的表中读取没有问题。我的设置是使用Glassfish 3.1.1的Netbeans 7.1，EclipseLink是我的持久性提供程序。我有一个非常简单的场景，我只想测试将人名和年龄写入数据库并使id自动递增。它是一个MySql数据库，包含字段：Id，FirstName和Age。继承我的代码：

Web servlet从html表单中获取名称和年龄：

    @Override
protected void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {

    String userPath = request.getServletPath();

    if(userPath.equals("/addUser")){

        //get request parameters from form
        String name = request.getParameter("name");
        String age = request.getParameter("age");

        //set request attributes to be used by forwarded page
        request.setAttribute("name", name);
        request.setAttribute("age", age);

        //create manager class to add person to database
        Manager manager = new Manager();
        manager.addPerson(name, age);

        userPath = "/result";
    }

    // use RequestDispatcher to forward request internally
    String url = "/WEB-INF/view" + userPath + ".jsp";

    try {
        request.getRequestDispatcher(url).forward(request, response);
    } catch (Exception ex) {
        ex.printStackTrace();
    }
}

接受名称和年龄的经理类，创建人物对象并保留它。

public class Manager {

private static final String PERSISTENCE_UNIT_NAME = "FormPU";
private static EntityManagerFactory factory;

public Manager() {
}

public void addPerson(String name, String age) {

    factory = Persistence.createEntityManagerFactory(PERSISTENCE_UNIT_NAME);
    EntityManager em = factory.createEntityManager();

    Persons persons = new Persons();
    persons.setName(name);
    persons.setAge(age);

    em.getTransaction().begin();
    em.persist(persons);
    em.getTransaction().commit();
    em.close();
}
}

人物实体类：

    /**
    *
    * @author esmiala
    */
    @Entity
    @Table(name = "persons")
    @XmlRootElement
    @NamedQueries({
      @NamedQuery(name = "Persons.findAll", query = "SELECT p FROM Persons p"),
      @NamedQuery(name = "Persons.findById", query = "SELECT p FROM Persons p WHERE  
      p.id = :id"),
      @NamedQuery(name = "Persons.findByFirstName", query = "SELECT p FROM Persons p 
      WHERE p.firstName = :firstName"),
      @NamedQuery(name = "Persons.findByAge", query = "SELECT p FROM Persons p WHERE 
      p.age = :age")})
public class Persons implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Basic(optional = false)
@NotNull
@Column(name = "Id")
private Integer id;
@Basic(optional = false)
@NotNull
@Size(min = 1, max = 255)
@Column(name = "FirstName")
private String firstName;
@Basic(optional = false)
@NotNull
@Size(min = 1, max = 255)
@Column(name = "Age")
private String age;

public Persons() {
}

public Persons(Integer id) {
    this.id = id;
}

public Persons(Integer id, String firstName, String age) {
    this.id = id;
    this.firstName = firstName;
    this.age = age;
}

public Integer getId() {
    return id;
}

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

public String getFirstName() {
    return firstName;
}

public void setFirstName(String firstName) {
    this.firstName = firstName;
}

public String getAge() {
    return age;
}

public void setAge(String age) {
    this.age = age;
}

@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 Persons)) {
        return false;
    }
    Persons other = (Persons) 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 "entity.Persons[ id=" + id + " ]";
}

}

的persistence.xml

    <?xml version="1.0" encoding="UTF-8"?>
    <persistence version="2.0" 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_2_0.xsd">
     <persistence-unit name="FormPU" transaction-type="JTA">
     <provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
     <jta-data-source>jdbc/form</jta-data-source>
     <exclude-unlisted-classes>false</exclude-unlisted-classes>
    <properties/>
    </persistence-unit>
    </persistence>

注意：我也尝试将exclude-unlisted-classes标记设置为true并单独列出该类，但这也不起作用。

例外：

    WARNING: StandardWrapperValve[Controller]: PWC1406: Servlet.service() for servlet  
    Controller threw exception
    java.lang.IllegalArgumentException: Object: entity.persons[ id=null ] is not a   
    known entity type.
atorg.eclipse.persistence.internal.sessions.UnitOfWorkImpl.registerNewObject
    ForPersist(UnitOfWorkImpl.java:4141)
atorg.eclipse.persistence.internal.jpa.EntityManagerImpl.   
    persist(EntityManagerImpl.java:368)
at manager.Manager.addPerson(Manager.java:36)
at controller.Controller.doPost(Controller.java:70)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:754)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:847)
at org.apache.catalina.core.StandardWrapper.service(StandardWrapper.java:1523)

......等等。任何帮助将不胜感激！

Answer 1

<exclude-unlisted-classes>无法正常工作 - persistence.xml中此元素的存在会禁用@Entity类的自动发现，无论其内部是什么。

此外，@Entity(name="persons")可能不是您想要的，请改用@Entity @Table (name="persons")。

Answer 2

所以你说你可以读得很好，但是在持续新实例时会出错？

您可以更新您阅读的对象吗？

您似乎遇到了某种类加载器问题。不知何故，你在类路径上有两次类，或者有两个不同的类加载器。您传递给的对象持久化来自与JPA正在使用的不同的类加载器。您可以检查读取内容的类加载器，以及持久化对象以查看它们之间的差异。

您是否已重新部署应用程序或已部署？如果正确关闭/重新启动服务器，它是否有效。确保在重新部署之前关闭旧的EntityManagerFactory。

Answer 3

关于具体问题，请尝试查看此link是否有帮助。

无论如何，实例化EntityManager的方式不是线程安全的。

您可以看到here为什么。或者，更好的是，您可以使用NetBeans的向导从实体类创建JPA控制器类，并查看它如何注入EntityManager：

@PersistenceContext
private EntityManager em;

另请参阅控制器类（相当于Manager POJO）具有Stateless注释。这是因为您可以安全地在其生命周期由Web容器管理的对象中注入EJB（在本例中为EntityManager）（有关访问Enterprise Bean的更多参考，请参阅here。）

对象：entity.ENTITY [id = null]不是已知的实体类型

3 个答案: