EntityManager查询返回一个奇怪的无法访问的对象

时间:2013-05-06 10:57:26

标签: java hibernate entity entitymanager hibernate-entitymanager

我有2个实体通过连接注释连接,一切正常,但查询结果很奇怪。

所以我让这堂课说 Cat ,另一堂课说 Home 。因此,如果我执行类 Cat 的命名查询,我希望使用表主页中的查询结果填充属性 Cat.home

我正在以这种方式执行查询:

List<Cat> a = (List<Cat>) em.createNamedQuery("Cat.findHome")
            .setParameter("catName", catName)
            .setParameter("houseKey", houseKey).getResultList();

我得到的结果是:

a = ArrayList<E>
       elementData= Object[10] (id=22688)
           [0] = Object[2] (id=22692)
              [0] = Cat (id=22692)
              [1] = Home (id=22692)
           [1] = null
           [2] = null
           [3] = null
           [4] = null
           [5] = null
           ...
           [9] = null

所以我的问题是如何访问这两个对象 Cat Home ,为什么我从实体管理器而不是单个对象获得此结果 Cat 主页位于 Cat.home

实体:

Cat.java - 猫实体

//imports here

@Entity
@Table(name="CAT")
@NamedQuery(name="cat.findHome", 
        query="from Cat a join a.home p where a.name = :catName and p.housekey like :houseKey")
public class Cat implements Serializable{

private static final long serialVersionUID = 1L;
private String catkey;
private String name;

public List<Home> key; // a cat can have many homes

/**
 * @param catkey the catkey to set
 */
public void setCatkey(String catkey) {
    this.catkey = catkey;
}

/**
 * @return the catkey
 */
@Id
@Column(name="K_CAT")
public String getCatkey() {
    return catkey;
}

/**
 * @return the name
 */
@Column(name="NAME")
public String getName() {
    return name;
}

/**
 * @param name the name to set
 */
public void setName(String name) {
    this.name = name;
}

/**
 * @param home the home to set
 */
public void setHome(List<Home> home) {
    this.home = home;
}

/**
 * @return the home
 */
@OneToMany(cascade=CascadeType.ALL, fetch=FetchType.LAZY, mappedBy="cat")
@Filter(name="contrFilter",condition = "K_CAT=:kcathome")
public List<Home> getHome() {
        return home;
    }
}

Home.java - 家庭实体

   @Entity
@Table(name="HOME")
public class Home implements Serializable{

    private static final long serialVersionUID = 1L;

private Sting housekey;
private Cat cat;

/**
 * @param housekey the housekey to set
 */
public void setHousekey(String housekey) {
    this.housekey = housekey;
}

/**
 * @return the housekey
 */
@Id
@Column(name="K_HOME")
public String getHousekey(){
    return housekey;
}

/**
 * @param cat the cat to set
 */
public void setCat(Cat cat) {
    this.cat = cat;
}

/**
 * @return the cat
 */
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "K_CAT_HOME", updatable=false, insertable=false)
public Cat getCat() {
        return cat;
    }
}

提前致谢!

1 个答案:

答案 0 :(得分:1)

select部分添加到命名查询

 query="select a from Cat a where a.name = :catName and a.home.houseKey like :houseKey")