如何插入使用Hibernate 4.1.10.Final?(How to insert using

2019-10-18 05:43发布

我使用休眠4.1.10.Final将数据插入到数据库中,但它抛出以下异常,

我有三个表,开发和地址。

发展具有地址的对象本身。

INFO: HCANN000001: Hibernate Commons Annotations {4.0.1.Final}
INFO: HHH000412: Hibernate Core {4.1.10.Final}
INFO: HHH000206: hibernate.properties not found
INFO: HHH000021: Bytecode provider name : javassist
INFO: HHH000043: Configuring from resource: /hibernate.cfg.xml
INFO: HHH000040: Configuration resource: /hibernate.cfg.xml
INFO: HHH000041: Configured SessionFactory: null
SEVERE: java.lang.NullPointerException
    at org.hibernate.mapping.PersistentClass.createPrimaryKey(PersistentClass.java:327)
    at org.hibernate.cfg.CreateKeySecondPass.doSecondPass(CreateKeySecondPass.java:48)
    at org.hibernate.cfg.Configuration.processSecondPassesOfType(Configuration.java:1386)
    at org.hibernate.cfg.Configuration.secondPassCompile(Configuration.java:1341)
    at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1730)
    at com.myproject.util.HibernateUtil.configureSessionFactory(HibernateUtil.java:23)
    at com.myproject.util.HibernateUtil.getSessionFactory(HibernateUtil.java:34)
    at com.myproject.model.ConstructionModel.addDevelopment(ConstructionModel.java:185)
    at com.myproject.controller.Construction.addDevelopment(Construction.java:135)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:601)
      ......

我的代码

development.java

@Entity
@Table(name="development")
public class Development implements Serializable{

    private int id;
    private Address address;

    public Development(){

    }
    @Id
    @GeneratedValue
    @Column(name="id")
    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }

    @OneToOne(cascade = CascadeType.ALL)
    @PrimaryKeyJoinColumn 
    public Address getAddress() {
        return this.address;
    }

    public void setAddress(Address address) {
        this.address = address;
    }


}

address.java

@Entity
@Table(name="address")
public class Address implements Serializable{

    private int id;
    private String unit;

    public Address(){

    }


    @Id
    @GeneratedValue
    @OneToOne(mappedBy="address")
    @JoinColumn(name="addid")
    public int getId() {
        return id;
    }


    @Column (name="unit")
    public String getUnit() {
        return unit;
    }

    public void setId(int id) {
        this.id = id;
    }


    public void setUnit(String unit) {
        this.unit = unit;
    }


}

Answer 1:

删除@GeneratedValue @OneToOne(mappedBy="development") @JoinColumn(name="Developer")@GeneratedValue @OneToOne(mappedBy="address") @JoinColumn(name="addid")从开发者和地址的标识字段。



文章来源: How to insert using hibernate 4.1.10.Final?