Hibernate:@JoinTable

Question

我已经理解了hibernate中的@joincolumn。现在我已经从@JoinTable开始了。以下是我的POJO

@Entity
@Table(name = "person")
public class Person {

    @Id
    @GeneratedValue
    @Column(name = "person_id")
    private long person_id;


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


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


    @OneToOne(cascade=CascadeType.ALL,fetch = FetchType.EAGER)  
        @JoinTable(name="person_phone",  
        joinColumns={@JoinColumn(name="person_id", referencedColumnName="person_id")},  
        inverseJoinColumns={@JoinColumn(name="phone_id", referencedColumnName="phone_id")})  
    private Phone phone;

        //Getters ande Setters

}


@Entity
@Table(name = "phone")
public class Phone{

    @Id
    @GeneratedValue
    @Column(name = "phone_id")
    private long phone_id;


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


        //Getters ande Setters


}

在我的控制器中，我完成了以下操作

 @RequestMapping(value = "/persons", method = RequestMethod.GET)
    public List<person> listPersons() {     

        StringBuilder sql = new StringBuilder();
        sql.append("SELECT * FROM person);  
        Query query = getSession().createSQLQuery(sql.toString()).addEntity(Person.class);
        return query.list();
    }

当我以Select*from person执行时，出现以下错误

o.h.e.jdbc.spi错误。SqlExceptionHelper-在此结果集中找不到列名phone_id。

我想实现的是，当我从人员表中检索记录时，我也希望相应的电话详细信息。

Answer 1

@Entity
@Table(name = "person")
public class Person implements java.io.Serializable {

    private long personId;  
    private String name;    
    private String address;
    private Phone phone;    

    public Person () {
    }

    @Id
    @GeneratedValue
    @Column(name = "person_id") 
    public BigDecimal getPersonId() {
        return this.personId;
    }

    @Column(name = "name")  
    public BigDecimal getName() {
        return this.name;
    }

    @Column(name = "Address")
    public BigDecimal getAddress() {
        return this.address;
    }

    @OneToOne(cascade=CascadeType.ALL,fetch = FetchType.EAGER)  
        @JoinTable(name="person_phone",  
        joinColumns={@JoinColumn(name="person_id", referencedColumnName="person_id")},  
        inverseJoinColumns={@JoinColumn(name="phone_id", referencedColumnName="phone_id")})  
    public Phone getPhone() {
        return this.phone;
    }
    // setter function
}

@Entity
@Table(name = "person_phone")
public class Phone{

    private long personId;
    private String name;

    @Id
    @GeneratedValue
    @Column(name = "person_id")
    public long getPersonId() {
        return this.personId;
    }

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

    //setter function
}

Answer 2

选择Hibernate createQuery方法而不是createSqlQuery来匹配实体更准确。试试下面；

    StringBuilder sql = new StringBuilder();
    sql.append("FROM Person");  //using hibernate query
    Query query = getSession().createQuery(sql.toString()); //Hibernate query so use createQuery not createSqlQuery
    return query.list();