Question

我使用Spring 4.2.5和org.hibernate 5.1.0。

当我使用entityManager.persist(user)保存user时，它无法保存到数据库，也没有错误抛出。

但如果我添加entityManager.getTransaction().begin();和entityManager.getTransaction().commit();，那就有效了。

下面的代码

@Service
@Transactional
public class UserTestService {

    @PersistenceUnit
    private EntityManagerFactory entityManagerFactory;


    public void addUser(User user){
        EntityManager entityManager = entityManagerFactory.createEntityManager();
        //entityManager.getTransaction().begin();
        entityManager.persist(user);
        //entityManager.getTransaction().commit();
    }
}

我的配置

@Configuration
@EnableTransactionManagement
public class JPAConfig  {

    @Autowired
    private DataSource dataSource;

    @Bean
    public LocalContainerEntityManagerFactoryBean entityManagerFactory() {
        LocalContainerEntityManagerFactoryBean entityManagerFactoryBean = new LocalContainerEntityManagerFactoryBean();
        entityManagerFactoryBean.setDataSource(dataSource);
        entityManagerFactoryBean.setJpaVendorAdapter(jpaVendorAdapter() );
        entityManagerFactoryBean.setPackagesToScan("zhihu.model");
        return entityManagerFactoryBean;
    }


    @Bean
    public JpaVendorAdapter jpaVendorAdapter() {
        HibernateJpaVendorAdapter adapter = new HibernateJpaVendorAdapter();
        adapter.setDatabase(Database.MYSQL);
        adapter.setShowSql(true);
        adapter.setGenerateDdl(false);
        adapter.setDatabasePlatform("org.hibernate.dialect.MySQL5Dialect");
        return adapter;
    }



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

        return transactionManager;
    }

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

我的模特

@Entity
@Table(name = "user")
public class User implements Serializable {

    @Id
    @GeneratedValue(strategy= GenerationType.IDENTITY)
    private long userID;

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

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

    public User(){

    }

    public User(long userID, String username, String password) {
        this.userID = userID;
        this.username = username;
        this.password = password;
    }

    public User(String username, String password) {
        this.username = username;
        this.password = password;
    }


    public long getUserID() {
        return userID;
    }

    public void setUserID(long userID) {
        this.userID = userID;
    }

    public String getUsername() {
        return username;
    }

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

    public String getPassword() {
        return password;
    }

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

的pom.xml

<dependencies>
        <dependency>
            <groupId>org.springframework</groupId>
            <artifactId>spring-jdbc</artifactId>
            <version>4.2.5.RELEASE</version>
        </dependency>
        <dependency>
            <groupId>org.springframework</groupId>
            <artifactId>spring-context</artifactId>
            <version>4.2.5.RELEASE</version>
        </dependency>
        <dependency>
            <groupId>mysql</groupId>
            <artifactId>mysql-connector-java</artifactId>
            <version>5.1.38</version>
        </dependency>
        <dependency>
            <groupId>javax.servlet</groupId>
            <artifactId>javax.servlet-api</artifactId>
            <version>3.1.0</version>
        </dependency>
        <dependency>
            <groupId>junit</groupId>
            <artifactId>junit</artifactId>
            <version>4.11</version>
            <scope>test</scope>
        </dependency>
        <dependency>
            <groupId>com.fasterxml.jackson.core</groupId>
            <artifactId>jackson-databind</artifactId>
            <version>2.7.3</version>
        </dependency>
        <dependency>
            <groupId>com.fasterxml.jackson.core</groupId>
            <artifactId>jackson-core</artifactId>
            <version>2.7.3</version>
        </dependency>
        <dependency>
            <groupId>com.fasterxml.jackson.core</groupId>
            <artifactId>jackson-annotations</artifactId>
            <version>2.7.3</version>
        </dependency>
        <dependency>
            <groupId>org.springframework.security</groupId>
            <artifactId>spring-security-web</artifactId>
            <version>4.0.4.RELEASE</version>
        </dependency>
        <dependency>
            <groupId>org.springframework.security</groupId>
            <artifactId>spring-security-config</artifactId>
            <version>4.0.4.RELEASE</version>
        </dependency>
        <dependency>
            <groupId>log4j</groupId>
            <artifactId>log4j</artifactId>
            <version>1.2.17</version>
        </dependency>

        <dependency>
            <groupId>org.hibernate</groupId>
            <artifactId>hibernate-entitymanager</artifactId>
            <version>5.1.0.Final</version>
        </dependency>
        <dependency>
            <groupId>org.springframework.data</groupId>
            <artifactId>spring-data-jpa</artifactId>
            <version>1.10.1.RELEASE</version>
        </dependency>
        <dependency>
            <groupId>org.springframework</groupId>
            <artifactId>spring-orm</artifactId>
            <version>4.2.5.RELEASE</version>
        </dependency>

    </dependencies>

我认为问题是Transactional。但是下面的调试日志。

2016-05-05 21:33:10 DEBUG TransactionImpl:51 - begin
2016-05-05 21:33:10 DEBUG TransactionImpl:62 - committing

更新

如果我添加entityManager.flush();导致javax.persistence.TransactionRequiredException: no transaction is in progress.有什么问题？

Answer 1

解决方案是注入EntityManager而不是EntityManagerFactory

Answer 2

提交将使数据库提交。对持久对象的更改将写入数据库。

当你有一个持久化对象而你在其上更改一个值时，它就会变脏，而hibernate需要将这些更改刷新到你的持久层。它可能会自动为您执行此操作，或者您可能需要手动执行此操作，具体取决于您的刷新模式（自动或手动）：）

最后你应该使用commit：transaction.commit（）会刷新会话，但它也会结束工作单元。

查看here了解更多信息。

Answer 3

如果您正在使用

entityManager.persist(user);

执行此语句后，更改将不会反映在数据库中。实体管理自动将这些更改刷新到您的持久层。

我们可以选择使用

手动清除它

entityManager.flush();

这应该可以解决您的问题

entityManager.persist不保存任何内容到数据库

3 个答案: