Question

我有基于Spring的应用程序，我需要让Hibernate工作。所以我受到了本教程的启发：A Guide to Hibernate with Spring 4我创建了java类（我想配置Hibernate xml-less）：

    package com.mycompany.cmms.web.config;



import java.util.Properties;

import javax.persistence.EntityManagerFactory;
import javax.sql.DataSource;

import org.springframework.context.annotation.Bean;
import org.springframework.context.annotation.Configuration;
import org.springframework.dao.annotation.PersistenceExceptionTranslationPostProcessor;
import org.springframework.jdbc.datasource.DriverManagerDataSource;
import org.springframework.orm.jpa.JpaTransactionManager;
import org.springframework.orm.jpa.JpaVendorAdapter;
import org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean;
import org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter;
import org.springframework.transaction.PlatformTransactionManager;
import org.springframework.transaction.annotation.EnableTransactionManagement;

@Configuration
@EnableTransactionManagement
public class PersistanceJPAConfig {


    @Bean
    public EntityManagerFactory entityManagerFactory() {
       LocalContainerEntityManagerFactoryBean em = new LocalContainerEntityManagerFactoryBean();
       em.setDataSource(dataSource());
       em.setPackagesToScan(new String[] { "com.vsb.cmms.domain" });

       JpaVendorAdapter vendorAdapter = new HibernateJpaVendorAdapter();
       em.setJpaVendorAdapter(vendorAdapter);
       em.setJpaProperties(additionalProperties());

       return em.getObject();
    }

    @Bean
    public DataSource dataSource(){
       DriverManagerDataSource dataSource = new DriverManagerDataSource();
       dataSource.setDriverClassName("org.postgresql.Driver");
       dataSource.setUrl("jdbc:postgresql://localhost:5432/cmms");
       dataSource.setUsername( "postgres" );
       dataSource.setPassword( "xxx" );
       return dataSource;
    }



    @Bean
    public PlatformTransactionManager transactionManager(EntityManagerFactory emf){
       JpaTransactionManager transactionManager = new JpaTransactionManager();
       transactionManager.setEntityManagerFactory(emf);

       return transactionManager;
    }

    @Bean
    public PersistenceExceptionTranslationPostProcessor exceptionTranslation(){
       return new PersistenceExceptionTranslationPostProcessor();
    }

    Properties additionalProperties() {
       Properties properties = new Properties();
       properties.setProperty("hibernate.hbm2ddl.auto", "create-drop");
       properties.setProperty("hibernate.dialect", "org.hibernate.dialect.PostgreSQLDialect");
       properties.setProperty("hibernate.query.factory_class", "org.hibernate.hql.internal.classic.ClassicQueryTranslatorFactory");
       return properties;
    }
}

但是当我构建并启动我的应用程序时，我得到了这个错误：

org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'transactionManager' defined in com.vsb.cmms.web.config.PersistanceJPAConfig: Invocation of init method failed; nested exception is java.lang.IllegalArgumentException: 'entityManagerFactory' or 'persistenceUnitName' is required
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1583)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:545)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory.java:482)
at org.springframework.beans.factory.support.AbstractBeanFactory$1.getObject(AbstractBeanFactory.java:306)
at org.springframework.beans.factory.support.DefaultSingletonBeanRegistry.getSingleton(DefaultSingletonBeanRegistry.java:230)
at org.springframework.beans.factory.support.AbstractBeanFactory.doGetBean(AbstractBeanFactory.java:302)
at org.springframework.beans.factory.support.AbstractBeanFactory.getBean(AbstractBeanFactory.java:197)
at org.springframework.beans.factory.support.DefaultListableBeanFactory.preInstantiateSingletons(DefaultListableBeanFactory.java:751)
at org.springframework.context.support.AbstractApplicationContext.finishBeanFactoryInitialization(AbstractApplicationContext.java:861)
at org.springframework.context.support.AbstractApplicationContext.refresh(AbstractApplicationContext.java:541)
at org.springframework.web.context.ContextLoader.configureAndRefreshWebApplicationContext(ContextLoader.java:444)
at org.springframework.web.context.ContextLoader.initWebApplicationContext(ContextLoader.java:326)
at org.springframework.web.context.ContextLoaderListener.contextInitialized(ContextLoaderListener.java:107)
at org.apache.catalina.core.StandardContext.listenerStart(StandardContext.java:4717)
at org.apache.catalina.core.StandardContext.startInternal(StandardContext.java:5179)
at org.apache.catalina.util.LifecycleBase.start(LifecycleBase.java:183)
at org.apache.catalina.core.ContainerBase$StartChild.call(ContainerBase.java:1404)
at org.apache.catalina.core.ContainerBase$StartChild.call(ContainerBase.java:1394)
at java.util.concurrent.FutureTask.run(FutureTask.java:266)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1142)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:617)
at java.lang.Thread.run(Thread.java:745)
Caused by: java.lang.IllegalArgumentException: 'entityManagerFactory' or 'persistenceUnitName' is required
at org.springframework.orm.jpa.JpaTransactionManager.afterPropertiesSet(JpaTransactionManager.java:304)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1642)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1579)
... 21 more

错误是说'entityManagerFactory'或'persistenceUnitName'是必需的。但实体经理工厂应该通过entityManagerFactory（）提供。当我对这个bean进行debbuged时，我注意到em.getObject（）返回null并且可能导致错误，尽管我不知道为什么会发生这种情况。我正在使用Spring 4和Hibernate 5.2.3.Final。

Answer 1

您的isAuth(): Promise<Status> { const url = "/account/check"; return this.http.get(url) .toPromise() .then(response=>return response.json().status as Status) .catch(this.handleError); }方法确实不应该调用entityManagerFactory()，而应该将其写为：

#getObject()

Spring处理工厂bean自动提供@Bean public LocalContainerEntityManagerFactoryBean entityManagerFactory() { LocalContainerEntityManagerFactoryBean em = new LocalContainerEntityManagerFactoryBean(); em.setDataSource(dataSource()); em.setPackagesToScan(new String[] { "com.vsb.cmms.domain" }); JpaVendorAdapter vendorAdapter = new HibernateJpaVendorAdapter(); em.setJpaVendorAdapter(vendorAdapter); em.setJpaProperties(additionalProperties()); // just return the bean here return em; }个实例。

Answer 2

刚刚在互联网的深处发现...... 如果你想轻松控制会话（应用程序的会话或事务的会话），那么你可能希望看到this

请注意，代码在Hibernate 5.2版本

之前完美无缺

使用JPA configration创建Spring bean时出现空指针异常

2 个答案: