Question

在这里，我想在公司@One与许多人（员工）之间建立关系和员工@Many到一个（公司）和@one到员工（人）和地址（Persondetail）之间的一个关系。

我在执行“我的主要课程”时面临以下错误

**解决此错误并让我编码打印有关员工（人）的所有数据，包括其他两个表记录**

信息：HHH000400：使用方言：org.hibernate.dialect.MySQL5Dialect 线程“main”中的异常org.hibernate.MappingException：无法确定类型：Rajat.Practice1.Company，在表：Person，for 列：[org.hibernate.mapping.Column（company）] 在org.hibernate.mapping.SimpleValue.getType（SimpleValue.java:336）在org.hibernate.mapping.SimpleValue.isValid（SimpleValue.java:310）在org.hibernate.mapping.Property.isValid（Property.java:241）在org.hibernate.mapping.PersistentClass.validate（PersistentClass.java:496）在org.hibernate.mapping.RootClass.validate（RootClass.java:270）在org.hibernate.cfg.Configuration.validate（Configuration.java:1358）在org.hibernate.cfg.Configuration.buildSessionFactory（Configuration.java:1849）在org.hibernate.cfg.Configuration.buildSessionFactory（Configuration.java:1928）在Rajat.Practice1.Main.main（Main.java:21）

（（MAIN CLASS））

Array<T>

公司课程

T[]

员工（人）班

package Rajat.Practice1;

    import javax.imageio.spi.ServiceRegistry;

    import org.hibernate.Session;
    import org.hibernate.SessionFactory;
    import org.hibernate.Transaction;
    import org.hibernate.cfg.AnnotationConfiguration;
    import org.hibernate.cfg.Configuration;
    import org.hibernate.service.ServiceRegistryBuilder;
    import org.hibernate.tool.hbm2ddl.SchemaExport;

    public class Main {


        public static void main(String[] args) {

             Configuration cfg=new Configuration();  
                cfg.configure("hibernate.cfg.xml");  

                SessionFactory sf=cfg.buildSessionFactory();  
                Session session=sf.openSession();  
                Transaction tx=session.beginTransaction();  

                PersonDetail pd=new  PersonDetail();
                pd.setCity("jaipur");
                pd.setState("jaipur");

                Company c=new Company();
                c.setC_name("Protechgenie");


            Person p=new Person();
            p.setName("Rajat");

            Person p1=new Person();
            p1.setName("Dynamo");

            p.setCompany(c);
            p1.setCompany(c);



            p.setP_detail(pd);

            session.save(c);
            session.save(p);
            session.save(p1);

            session.getTransaction().commit();

        }

    }

员工地址（人员详细信息类）`

    package Rajat.Practice1;

    import java.util.List;

    import javax.persistence.Column;
    import javax.persistence.ElementCollection;
    import javax.persistence.Entity;
    import javax.persistence.FetchType;
    import javax.persistence.GeneratedValue;
    import javax.persistence.Id;
    import javax.persistence.OneToMany;
    import javax.persistence.Table;

    @Entity
    @Table(name="company")
    public class Company {

        @Id
        @GeneratedValue
        @Column(name="c_id")
        private int c_id;
        private String C_name;
        @ElementCollection(targetClass=Person.class)
        private List <Person> person;

        @OneToMany(targetEntity=Person.class, mappedBy="person" , cascade=javax.persistence.CascadeType.ALL, fetch=FetchType.EAGER)
        public List<Person> getPerson() {
            return person;
        }
        public void setPerson(List<Person> person) {
            this.person = person;
        }
        public int getC_id() {
            return c_id;
        }
        public void setC_id(int c_id) {
            this.c_id = c_id;
        }
        public String getC_name() {
            return C_name;
        }
        public void setC_name(String c_name) {
            C_name = c_name;
        }


    }

Answer 1

我尝试在下面逐一纠正解决问题的类定义

@Entity
public class Company {

    private int c_id;
    private String name;
    //@ElementCollection(targetClass=Person.class)

    private List<Person> person = new ArrayList<Person>();


    @OneToMany(mappedBy="company",fetch = FetchType.LAZY)
    public List<Person> getPerson() {
        return person;
    }
    public void setPerson(List<Person> person) {
        this.person = person;
    }
    @Id
    @GeneratedValue
    @Column(name="c_id")
    public int getc_id() {
        return c_id;
    }
    public void setc_id(int c_id) {
        this.c_id = c_id;
    }
    @Column(name="name")
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
}

人类

@Entity
public class Person {

    private int Pid;

    private String name;

    private Company company;

    private PersonDetail p_detail;

    @OneToOne ( cascade=javax.persistence.CascadeType.ALL, fetch=FetchType.EAGER  )
    @JoinColumn(name="pdetail_pk")
    public PersonDetail getP_detail() {
        return p_detail;
    }
    public void setP_detail(PersonDetail p_detail) {
        this.p_detail = p_detail;
    }

    @Id
    @GeneratedValue
    @Column(name="Pid_pk")
    public int getPid() {
        return Pid;
    }
    public void setPid(int pid) {
        Pid = pid;
    }

    @Column(name="name")
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }


    @ManyToOne(fetch=FetchType.LAZY)
    @JoinColumn(name="c_id", nullable = false)
    public Company getCompany() {
        return company;
    }

    public void setCompany(Company company) {
        this.company = company;
    }

}

PersonDetail类

@Entity
@Table(name="pdetail")
public class PersonDetail {

    @Id
    @GeneratedValue
    @Column(name="pdetail_pk")
    private int p_id;


    @Column(name="city")
    private String City;


    @Column(name="state")
    private String State;



    public String getCity() {
        return City;
    }
    public void setCity(String city) {
        City = city;
    }
    public String getState() {
        return State;
    }
    public void setState(String state) {
        State = state;
    }

}

hibernate中多对一关系出错

1 个答案: