在Hibernate中持久化LinkedList

Question

我试图坚持一个具有LinkedList属性的类,但似乎无法正确.这是我的代码和我的映射:

 import java.util.LinkedList;
 public class Stuff implements java.io.Serializable {
    private long id;
private LinkedList<Image> images;


public Stuff() {
}


public Stuff(long Id) {
    id = Id;
}

public long getId() {
    return id;
}

public void setId(long mealId) {
    id = mealId;
}


public LinkedList<Image> getNumberImages(int number) {
    assert (number >= 0);
    return (LinkedList<Image>) images.subList(0, number) ;
}

public LinkedList<Image> getImages() {
    return images;
}
    public LinkedList<Image> setImages(LinkedList<Image> images) {
    this.images = images;
}


public void addImage(Image image) {
    if (!images.contains(image)) {
        images.add(image);
    }
}

Hibernate映射:

<hibernate-mapping>
    <class name="data.Stuff" table="Stuff">
        <id name="id" type="long" access="field">
            <column name="ID" />
            <generator class="assigned" />
        </id>
        <list name="images" inverse="false" table="IMAGE" lazy="true" access="field">
            <key>
                <column name="ID" />
            </key>
            <list-index></list-index>
            <one-to-many class="data.Image" />
        </list>
    </class>
</hibernate-mapping>

似乎我可以像这样坚持Class Stuff的对象但是当我尝试恢复它们时会发生以下错误:

Hibernate: select stuff0_.ID as ID0_, stuff0_.NAME as NAME0_, meal0_.GROUPING as GROUPING0_ from MEAL meal0_
org.hibernate.PropertyAccessException: could not set a field value by reflection setter of data.Meal.images

Answer 1

通常,Hibernate将为集合提供自己的实现,因此您应该更喜欢特定实现的接口.它可能试图为图像分配不同类型的列表并失败.您必须将您的字段更改为List<Image>.

Answer 2

Hibernate(以及一般的JPA)使用其接口来持久化集合.该列表必须声明为a List,而不是a LinkedList.并且它不会加载LinkedList实例,因为Hibernate使用自己的List实现来实现脏检查,延迟加载等.

对接口进行编程是一种很好的做法,而不是通常对具体实现进行编程.在JPA实体中,它是强制性的.