Hibernate沒有找到映射文件：org.hibernate.UnknownEntityTypeException：找不到persist

Question

我使用hibernate 5並嘗試將對象保存到我的數據庫。但由於某種原因，我總是得到一個 Exception in thread "main" org.hibernate.MappingException: Unknown entity: model.database.Customer。

出於某種原因，Customer.hbm.xml沒有被發現。我真的不知道爲什麼。

Customer.hbm.xml：

<?xml version='1.0' encoding='utf-8'?> 
<!DOCTYPE hibernate-mapping PUBLIC 
"-//Hibernate/Hibernate Mapping DTD 3.0//EN" 
"http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd"> 
<hibernate-mapping> 
<class name="model.database.Customer" table="CUSTOMER"> 
    <id name="username" type="string"> 
     <column name="USERNAME" length="8" /> 
     <generator class="assigned"></generator> 
    </id> 
    <property name="password" type="string"> 
     <column name="PASSWORD" length="8" /> 
    </property> 
    <property name="lastname" type="string"> 
     <column name="LASTNAME" length="20" /> 
    </property> 
    <property name="firstname" type="string"> 
     <column name="FIRSTNAME" length="20" /> 
    </property> 
</class> 
</hibernate-mapping> 

Customer.java：

import javax.persistence.Column; 
import javax.persistence.Entity; 
import javax.persistence.Id; 
import javax.persistence.Table; 

@Entity 
@Table(name="CUSTOMER") 
public class Customer implements java.io.Serializable { 



    private static final long serialVersionUID = 1L; 

    @Id 
    @Column(name="username", nullable=false) 

    private String username; 

    @Column(name="firstname") 
    private String firstname; 

    @Column(name="lastname") 
    private String lastname; 

    @Column(name="password") 
    private String password; 

    public Customer() { 

    } 

    public Customer(String username) { 
     this.username = username; 
    } 

    public Customer(String username, String firstname, String lastname, String password) { 
     this.username = username; 
     this.firstname = firstname; 
     this.lastname = lastname; 
     this.password = password; 
    } 

    public String getUsername() { 
     return this.username; 
    } 

    public void setUsername(String username) { 
     this.username = username; 
    } 


    public String getFirstname() { 
     return this.firstname; 
    } 

    public void setFirstname(String firstname) { 
     this.firstname = firstname; 
    } 


    public String getLastname() { 
     return this.lastname; 
    } 

    public void setLastname(String lastname) { 
     this.lastname = lastname; 
    } 


    public String getPassword() { 
     return this.password; 
    } 

    public void setPassword(String password) { 
     this.password = password; 
    } 


} 

hibernate.cfg.xml中：

<?xml version="1.0" encoding="utf-8"?> 
<!DOCTYPE hibernate-configuration SYSTEM 
     "http://www.hibernate.org/dtd/hibernate-configuration-3.0.dtd"> 
<hibernate-configuration> 
    <session-factory> 
    <property name="hibernate.dialect"> 
     org.hibernate.dialect.PostgreSQL81Dialect 
    </property> 
    <property name="hibernate.connection.driver_class"> 
     org.postgresql.Driver 
    </property> 
    <property name="hibernate.connection.url"> 
     jdbc:postgresql://localhost:5432/postgres 
    </property> 
    <property name="hibernate.connection.username"> 
     postgres 
    </property> 
    <property name="hibernate.connection.password"> 
     postgres 
    </property> 
    <property name="hibernate.current_session_context_class"> 
     thread 
    </property> 

    <mapping resource="model/database/Customer.hbm.xml"/> 
    </session-factory> 
</hibernate-configuration> 

主要：

public class Main { 

    public static void main(String[] args) { 
     Customer cstm = new Customer("lebetyp", "peter", "ja", "ja"); 
     CustomerManager mngr = new CustomerManager(); 
     mngr.saveCustomer(cstm); 
    } 
} 

的HibernateUtil：

public class HibernateUtil { 

    private static final SessionFactory sessionFactory; 

    static { 
     try { 
      // Create the SessionFactory from hibernate.cfg.xml 
      Configuration configuration = new Configuration().configure(); 
      StandardServiceRegistryBuilder builder = new StandardServiceRegistryBuilder().applySettings(configuration.getProperties()); 
      sessionFactory = configuration.buildSessionFactory(builder.build()); 
      System.out.println("Initial SessionFactory creation"); 
     } catch (Throwable ex) { 
      System.out.println("Initial SessionFactory creation failed." + ex); 
      throw new ExceptionInInitializerError(ex); 
     } 
    } 

    public static SessionFactory getSessionFactory() { 
     return sessionFactory; 
    } 

} 

我怎樣才能擺脫這種例外的，使其工作。有沒有依賴？或者我做錯了什麼？

Answer 1

添加到您的hibernate.cfg.xml

刪除<mapping resource="model/database/Customer.hbm.xml"/>並添加

<mapping class="Customer"></mapping>. 

我希望你的客戶類有JPA註釋像@實體，@Id等在裏面。