Question

我试图通过Spring和Java Persistence API在我的数据库中持久化一个Test类型的对象

我的数据库中有一个Test表，我创建了相应的Entity Class：

   @Entity
   @Table(name = "test")
   @XmlRootElement
   @NamedQueries({
   @NamedQuery(name = "Test2.findAll", query = "SELECT t FROM Test t"),
   @NamedQuery(name = "Test2.findByTestId", query = "SELECT t FROM Test t WHERE t.testId = :testId"),
   @NamedQuery(name = "Test2.findByTestint", query = "SELECT t FROM Test t WHERE t.testint = :testint"),
   @NamedQuery(name = "Test2.findByText", query = "SELECT t FROM Test t WHERE t.text = :text")})

public class Test implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Basic(optional = false)
@Column(name = "testId")
private Integer testId;
@Column(name = "testint")
private Integer testint;
@Basic(optional = false)

@Temporal(TemporalType.TIMESTAMP)
private Date endtime;
@Column(name = "text")
private String text;

public Test() {
}

public Test(Integer testId) {
    this.testId = testId;
}

public Test(Integer testId, Date starttime, Date endtime) {
    this.testId = testId;
    this.starttime = starttime;
    this.endtime = endtime;
}

public Integer getTestId() {
    return testId;
}

public void setTestId(Integer testId) {
    this.testId = testId;
}

public Integer getTestint() {
    return testint;
}

public void setTestint(Integer testint) {
    this.testint = testint;
}



public String getText() {
    return text;
}

public void setText(String text) {
    this.text = text;
}
}

然后我创建了一个实现TestDao接口的Jpa类：

 @Repository("testDao")
public class JpaTestDao  implements TestDao {


public JpaTestDao(EntityManagerFactory emf) {
    this.emf = emf;
}
@PersistenceUnit
private EntityManagerFactory emf;// = null;
EntityManager em;

public EntityManager getEntityManager() {
    return emf.createEntityManager();
}
@Transactional
@Override
public void create(Test test) {
    em = null;
    try {
        em = getEntityManager();
        em.getTransaction().begin();
        em.persist(test);
        em.getTransaction().commit();
    } finally {
        if (em != null) {
            em.close();
        }
    }
}
}

TestDao界面：

public interface TestDao {

    public void create(Test t);
}

最后，我收到以下错误：

 Could not instantiate bean class [com.mycompany.jpa.JpaTestDao]: No default constructor   found; nested exception is java.lang.NoSuchMethodException: com.mycompany.jpa.JpaTestDao.

请你帮助我吗？我无法弄清楚出了什么问题？

Answer 1

因为你在JpaTestDao中添加了一个额外的构造函数，如下所示：

public JpaTestDao(EntityManagerFactory emf) {
    this.emf = emf;
}

Java将不再为您生成默认构造函数。您需要添加默认构造函数：

JpaTestDao() {}

这是非常标准的Java行为。在转向Spring之前，绝对建议在Java基础知识上更加扎实。或者，也许你只是犯了一个诚实的错误。：）

此外，正如其他人所提到的，您将需要注入EntityManager而不是注入EntityManagerFactory。然后，您需要使用以下命令注释EntityManager：

@PersistenceContext(unitName="greatUnitName")

您将在Spring Config中设置unitName。 JavaConfig示例如下：

@Bean
    public LocalContainerEntityManagerFactoryBean entityManagerFactory() throws SQLException
    {
        LocalContainerEntityManagerFactoryBean factory = new LocalContainerEntityManagerFactoryBean();
        factory.setDataSource(dataSource());
        factory.setJpaVendorAdapter(envConfig.jpaVendorAdapter());
        factory.setPackagesToScan("java.pkg.pkg1","java.pkg.pkg2");
            factory.setPersistenceUnitName("greatUnitName");

        return factory;
    }

    @Bean
    public EntityManager entityManager() throws SQLException
    {
        return entityManagerFactory().getObject().createEntityManager();
    }

    @Bean
    public PlatformTransactionManager transactionManager() throws SQLException
    {
        JpaTransactionManager jpaTransactionManager = new JpaTransactionManager();
        jpaTransactionManager.setEntityManagerFactory(entityManagerFactory().getObject());
        return jpaTransactionManager;
    }

    //... snip DataSource setup

错误：无法实例化bean类 - 找不到默认构造函数;

1 个答案: