java - hibernate : @JoinTable

Question

我在hibernate中理解了@joincolumn。现在我开始使用@JoinTable。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 Controller 中，我做了以下操作

 @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();
    }

当我执行 Select * from person 时。我收到以下错误

ERROR o.h.e.jdbc.spi.SqlExceptionHelper - The column name phone_id was not found in this ResultSet.

我想要实现的是，当我从 person 表中检索记录时，我也想要相应的电话详细信息。

Answer 1

选择Hibernate createQuery方法而不是createSqlQuery来匹配实体更准确。 请尝试以下；

    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();