本文介绍了休眠:@JoinTable的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我在休眠状态下了解了@joincolumn.现在，我从@JoinTable开始.以下是我的POJO
I have understood @joincolumn in hibernate. Now I have started with @JoinTable. Following is my 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，我已完成以下操作
In my controller, I have done the following
@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();
}
，当我执行时，从person中选择*.我收到以下错误
and when I execute as Select * from person . I am getting the following error
我想要实现的是，当我从 person 表中检索记录时，我还想要相应的电话详细信息.
What I want to achieve is , when I retrieve a record from person table I want the corresponding phone details also .
推荐答案
选择Hibernate createQuery方法而不是createSqlQuery来匹配实体更为准确.请尝试以下；
Choosing Hibernate createQuery method instead createSqlQuery to match entity is more accurate.Try below;
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();
这篇关于休眠:@JoinTable的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！