休眠查询。演员类

Question

有hql查询

List<Developer> developers = session.createQuery("from Developer d " +
              "left join ProjectDeveloper pd on pd.developer.id = d.id " +
              "where pd.project.id = " + project.getId()).getResultList();

它可以工作并获取对象列表，但不是开发人员。

实际上，带有i get 的对象看起来像一个数组列表，其中包括对象数组，其中包括开发人员class 对象和ProjectDeveloper (mtm class)。不幸的是，我无法在图像中放置 link，但架构看起来像：

developers = {ArrayList@4889} size = 4 
 \/ 0 = {object[2]4897} //what does this do in ArrayList<**Developer**>>??
    -> 0 = {Developer@4901} //This is exactly the class I expect in ArrayList
    -> 1 = {ProjectDeveloper}//?????
 -> 1 = {object[2]4898}

开发者class：

@Table(name = "developers")
@Entity
public class Developer implements GenerallyTable{

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id")
    private long id;
    @Column(name = "first_name")
    private String name;     
    @Column(name = "age")
    private int age;     
    @Column(name = "sex")
    private boolean sex;    
    @Column(name = "salary")
    private BigDecimal salary;
//...getters + setters
}

这怎么不可能（Developer.class 如何转换为具有未知结构的数组），以及如何准确获取 Developers 的 ArrayList？

Answer 1

尝试使用 createQuery 定义的类型 https://docs.jboss.org/hibernate/orm/5.2/javadocs/org/hibernate/Session.html#createQuery-java.lang.String-java.lang.Class-

List<Developer> developers = session.createQuery("select d from Developer d " +
                    "left join ProjectDeveloper pd on pd.developer.id = d.id " +
                    "where pd.project.id = " + project.getId(), Developer.class)
                    .getResultList();

你还应该使用 setParameter 方法

List<Developer> developers = session.createQuery("select d from Developer d " +
                    "left join ProjectDeveloper pd on pd.developer.id = d.id " +
                    "where pd.project.id = :proj_id", Developer.class)
                    .setParameter("proj_id", project.getId())
                    .getResultList();

休眠查询。演员 类

Hibernate query. Cast classes

java

hibernate

hql

休眠查询。演员类