[英]Hibernate mapping exception
我在嘗試將Employee
對象保存到數據庫時遇到以下錯誤:
org.hibernate.MappingException: Unknown entity: com.example.common.Employee
at org.hibernate.impl.SessionFactoryImpl.getEntityPersister(SessionFactoryImpl.java:597)
at org.hibernate.impl.SessionImpl.getEntityPersister(SessionImpl.java:1339)
與休眠有關的 pom.xml中的依賴項 [我的是Spring-Hibernate項目] **
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate</artifactId>
<version>3.2.7.ga</version>
</dependency>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-entitymanager</artifactId>
<version>4.3.4.Final</version>
</dependency>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-validator</artifactId>
<version>5.1.0.Final</version>
</dependency>
<dependency>
<groupId>org.springframework</groupId>
<artifactId>spring-orm</artifactId>
<version>3.2.2.RELEASE</version>
</dependency>
我曾經在映射XML文件中定義Hibernate映射。 在這個例子中,我嘗試使用注釋。
Employee.java [僅代碼的開頭部分]
我也包括導入語句
import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.Id;
import javax.persistence.ManyToMany;
import javax.persistence.OneToMany;
import javax.persistence.Table;
import javax.validation.Valid;
import javax.validation.constraints.NotNull;
import javax.validation.constraints.Size;
import org.hibernate.validator.constraints.NotEmpty;
@Entity
@Table(name = "employees")
public class Employee implements Serializable {
private static final long serialVersionUID = 5468763051360122059L;
@Id
@Column(name = "employee_guid", length = 36)
@NotNull
@Size(min = 36, max = 36)
private String id;
@Column(name = "is_active", nullable = false)
private boolean isActive;
@ManyToMany(mappedBy = "employees", fetch = FetchType.LAZY, cascade = {
CascadeType.PERSIST, CascadeType.MERGE, CascadeType.REFRESH, CascadeType.DETACH
})
我也嘗試將@Enitiy
更改為@org.hibernate.annotations.Entity
,但是沒有用。
請給我一些提示,指出我做錯了什么。 如果您需要更多代碼,請告訴我。
編輯:
讓我知道如何在這里指定Employee Entity bean?
<bean id="MyHibernateSessionFactory"
class="org.springframework.orm.hibernate3.LocalSessionFactoryBean">
<property name="dataSource" ref="dataSource" />
<property name="mappingResources">
<!-- if there is a mapping file like Employee.hbm.xml i can refer here, but i manage thru annotations -->
</property>
<property name="hibernateProperties">
<props>
<prop key="hibernate.show_sql">true</prop>
<prop key="hibernate.hbm2ddl.auto">create</prop>
<prop key="hibernate.connection.autocommit">false</prop>
<prop key="hibernate.connection.isolation">2</prop>
<prop key="hibernate.transaction.auto_close_session">true</prop>
<prop key="hibernate.current_session_context_class">managed</prop>
<prop key="hibernate.transaction.flush_before_completion">true</prop>
<prop key="hibernate.dialect">org.hibernate.dialect.H2Dialect</prop>
</props>
</property>
</bean>
我正在嘗試這個
<property name="annotatedClasses">
<list>
<value>com.example.Employee</value>
</list>
</property>
也已更改: class="org.springframework.orm.hibernate3.annotation.AnnotationSessionFactoryBean">
更新:
除非共享更多代碼,否則這個問題是不完整的。 因此,現在關閉此問題,並以以下建議作為答案,盡管這並不是我一直在尋找的確切答案。 如果有機會,我將重新開放。 謝謝大家
除了注釋,您還需要讓Hibernate知道您的實體類。 我認為有兩種方法可以做到這一點:
1)將類添加到您的hibernate.cfg.xml文件中:
<hibernate-configuration>
<session-factory>
<mapping class="com.example.common.Employee" />
</session-factory>
</hibernate-configuration>
2)在代碼中配置Hibernate:
import org.hibernate.cfg.Configuration;
import org.hibernate.service.ServiceRegistry;
private static SessionFactory buildSessionFactory() {
Configuration configuration = new Configuration();
configuration.configure();
configuration.addAnnotatedClass(Employee.class);
ServiceRegistry serviceRegistry = new ServiceRegistryBuilder().applySettings(configuration.getProperties()).buildServiceRegistry();
return configuration.buildSessionFactory(serviceRegistry);
}
休眠實體必須定義一個無參數的構造函數。
資料來源: http : //coding.tocea.com/java/java-frameworks/hibernate/hibernate-entity-does-not-define-no-argument-constructor/
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