Hibernate:为什么FetchType.LAZY-annotated集合属性急切加载?

inh*_*dle 8 java hibernate jpa

我试图实现简单的一对多关联.在使用调试模式检查项目对象后,我发现已加载List <Bid>出价.但是List <Bid>出价属性使用FetchType.LAZY进行批注.一些书籍和网页声称FetchType.LAZY是JPA提供商接受或拒绝的提示.但我想知道JPA提供商在什么条件下忽略了FetchType.LAZY.先感谢您.

@Entity
@Table(name = "ITEM")
public class Item implements Serializable {

    @Id
    private Long id = null;

    private String name;
    @ManyToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "SELLER_ID", nullable = false)
    private User seller;

    @OneToMany(mappedBy = "item", fetch = FetchType.LAZY)
    private List<Bid> bids;

    /**
     * No-arg constructor for JavaBean tools.
     */
    public Item() {}

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public User getSeller() {
        return seller;
    }

    public void setSeller(User seller) {
        this.seller = seller;
    }

    @Override
    public String toString() {
        return "Item{" +
                "id=" + id +
                ", name='" + name + '\'' +
                ", seller=" + seller +
                ", bids=" + bids +
                '}';
    }
}
@Entity
@Table(name = "BID")
public class Bid implements Serializable {

    @Id @GeneratedValue
    @Column(name = "BID_ID")
    private Long id = null;

    @ManyToOne
    @JoinColumn(name = "ITEM_ID", nullable = false, updatable = false, insertable = false)
    private Item item;

    @ManyToOne
    @JoinColumn(name = "BIDDER_ID", nullable = false, updatable = false)
    private User bidder;

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public Item getItem() {
        return item;
    }

    public void setItem(Item item) {
        this.item = item;
    }

    public User getBidder() {
        return bidder;
    }

    public void setBidder(User bidder) {
        this.bidder = bidder;
    }

    @Override
    public String toString() {
        return "Bid{" +
                "id=" + id +
                ", bidder=" + bidder +
                '}';
    }
}
private static void itemSeller(EntityManager em) {

        Item item = em.find(Item.class, 1L);
        System.out.println("hello");

    }

编辑:我在声明中提出了断点System.out.println("hello").我检查了物品对象.看到图片: 在此输入图像描述

JB *_*zet 17

通过检查调试器中的对象,您要求它调用列表的方法来显示其大小,内容等.当然,这会懒惰地初始化列表.

对于您的toString()方法也是如此,它隐含地遍历列表以进行打印.

  • 看到你的回答,我觉得我的问题很愚蠢。因为太丢人了,我考虑过删除这篇文章。现在我对赞成票的数量感到惊讶,我很高兴即使我的愚蠢问题对其他人也派上了用场。再次感谢您的回答。 (3认同)
  • @inherithandle,没有必要感到羞耻,只有当我们尝试时才会学习,我在谷歌上输入了同样的担忧,并登陆了你的线程,看看一个问题如何帮助其他人。感谢您的询问 (2认同)