java - Hibernate - 如何在一个查询中获取对象列表的属性

Question

我有 Persistent 类，引用了许多具有 LAZY fetch 类型的集合，例如

@Entity
@Table(name = "TABLE")
public class Table implements Serializable {
    ....


    @ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
    @JoinColumn(name = "LIST1", nullable = true)
    private ArrayList list1;

    @ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
    @JoinColumn(name = "LIST2", nullable = true)
    private ArrayList list2;

    @ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
    @JoinColumn(name = "LIST1", nullable = true)
    private ArrayList list1;

    @ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
    @JoinColumn(name = "LIST2", nullable = true)
    private ArrayList list2;

....

}

我已经通过 Hibernate 实例化了一些 TABLE 类型的对象，将其添加到表列表中，现在想要为该列表中的所有对象获取该集合之一(例如 list2)。

for(Table table:tables){
     result=table.list2;
     ....
}

但是这样 Hibernate 将生成单独的 SQL 查询序列。 hibernate可以在一次查询中获取集合中所有对象的list2吗？ (重要的是不要创建新的 Table 类实例，而是修改已经存在的对象)

Answer 1

来自the reference manual :

A "fetch" join allows associations or collections of values to be initialized along with their parent objects using a single select. This is particularly useful in the case of a collection. It effectively overrides the outer join and lazy declarations of the mapping file for associations and collections.

select t from Table t
left join fetch t.list2
where ...