Jpa 无法显示链接关联中的值

我有一个带有JPA和Thymeleaf的Spring Boot应用程序。我无法在

th

模板中显示与

提名.提名人.雇员.姓名

等效的内容

这里需要注意的是，如果我使用提名控制器来表示提名人，则员工姓名将通过

${namegate.employeepretyname}

显示在其视图中

以下是POJO代码：

//I am working with a NominationDraft Type instance of this class
public abstract class Nomination {
  @OneToMany
  private List<Nominee> nominees;

  public List<Nominee> getNominees() {
      return nominees;
  }
}

public class Nominee {

  @Id
  @GeneratedValue(strategy = GenerationType.AUTO)
  private long id;

  @ManyToOne
  private Nomination nomination;

  @ManyToOne
  private Employee employee;

  public Employee getEmployee() {
      return employee;
  }

  public String getEmployeePrettyName() { return employee.getPrettyName(); }

以下是有问题的控制器方法：

@RequestMapping(value="/nomination/{id}")
public String nomination(@PathVariable("id") Long id, Model model) {
    Nomination nomination = nominationService.findById(id);
    model.addAttribute("nomination", nomination);
    model.addAttribute("nominees", nomination.getNominees());
    return "nomination";
}

Hibernate: select nomination0_.id as id2_3_, nomination0_.user_id as user_id3_3_, nomination0_.nomination_type as nominati1_3_ from nomination nomination0_ where nomination0_.id=?
Hibernate: select user0_.id as id1_7_0_, user0_.employee_id as employee2_7_0_, employee1_.id as id1_0_1_, employee1_.email as email2_0_1_, employee1_.first_name as first_na3_0_1_, employee1_.last_name as last_nam4_0_1_, employee1_.manager_id as manager_6_0_1_, employee1_.nnumber as nnumber5_0_1_, directrepo2_.manager_id as manager_6_0_2_, directrepo2_.id as id1_0_2_, directrepo2_.id as id1_0_3_, directrepo2_.email as email2_0_3_, directrepo2_.first_name as first_na3_0_3_, directrepo2_.last_name as last_nam4_0_3_, directrepo2_.manager_id as manager_6_0_3_, directrepo2_.nnumber as nnumber5_0_3_ from user user0_ left outer join employee employee1_ on user0_.employee_id=employee1_.id left outer join employee directrepo2_ on employee1_.id=directrepo2_.manager_id where user0_.id=?
Hibernate: select nominees0_.nomination_id as nominati1_3_0_, nominees0_.nominees_id as nominees2_4_0_, nominee1_.id as id1_5_1_, nominee1_.employee_id as employee2_5_1_, nominee1_.nomination_id as nominati3_5_1_, employee2_.id as id1_0_2_, employee2_.email as email2_0_2_, employee2_.first_name as first_na3_0_2_, employee2_.last_name as last_nam4_0_2_, employee2_.manager_id as manager_6_0_2_, employee2_.nnumber as nnumber5_0_2_, employee3_.id as id1_0_3_, employee3_.email as email2_0_3_, employee3_.first_name as first_na3_0_3_, employee3_.last_name as last_nam4_0_3_, employee3_.manager_id as manager_6_0_3_, employee3_.nnumber as nnumber5_0_3_, nomination4_.id as id2_3_4_, nomination4_.user_id as user_id3_3_4_, nomination4_.nomination_type as nominati1_3_4_, user5_.id as id1_7_5_, user5_.employee_id as employee2_7_5_, employee6_.id as id1_0_6_, employee6_.email as email2_0_6_, employee6_.first_name as first_na3_0_6_, employee6_.last_name as last_nam4_0_6_, employee6_.manager_id as manager_6_0_6_, employee6_.nnumber as nnumber5_0_6_ from nomination_nominees nominees0_ inner join nominee nominee1_ on nominees0_.nominees_id=nominee1_.id left outer join employee employee2_ on nominee1_.employee_id=employee2_.id left outer join employee employee3_ on employee2_.manager_id=employee3_.id left outer join nomination nomination4_ on nominee1_.nomination_id=nomination4_.id left outer join user user5_ on nomination4_.user_id=user5_.id left outer join employee employee6_ on user5_.employee_id=employee6_.id where nominees0_.nomination_id=?

以下是视图文件：

<h1>Viewing a Nomination</h1>
<table>
<tr th:each="nominee : ${nominees}">
    <td th:text="${nominee.employeePrettyName}">Pretty Name</td>
</tr>
</table>

---

谢谢

如果一项提名有许多被提名人，你不应该为被提名人编制索引，以便能够找到你想要的人吗

如下所示：

namignment.namegate[i].employee.name

其中i是要显示的被提名人的索引

---

参考资料：

Fair point，但我认为thymeleaf旨在允许简单地迭代集合，而不知道要显示的每个对象的索引。这就是我使用

Hibernate: select nomination0_.id as id2_3_, nomination0_.user_id as user_id3_3_, nomination0_.nomination_type as nominati1_3_ from nomination nomination0_ where nomination0_.id=?
Hibernate: select user0_.id as id1_7_0_, user0_.employee_id as employee2_7_0_, employee1_.id as id1_0_1_, employee1_.email as email2_0_1_, employee1_.first_name as first_na3_0_1_, employee1_.last_name as last_nam4_0_1_, employee1_.manager_id as manager_6_0_1_, employee1_.nnumber as nnumber5_0_1_, directrepo2_.manager_id as manager_6_0_2_, directrepo2_.id as id1_0_2_, directrepo2_.id as id1_0_3_, directrepo2_.email as email2_0_3_, directrepo2_.first_name as first_na3_0_3_, directrepo2_.last_name as last_nam4_0_3_, directrepo2_.manager_id as manager_6_0_3_, directrepo2_.nnumber as nnumber5_0_3_ from user user0_ left outer join employee employee1_ on user0_.employee_id=employee1_.id left outer join employee directrepo2_ on employee1_.id=directrepo2_.manager_id where user0_.id=?
Hibernate: select nominees0_.nomination_id as nominati1_3_0_, nominees0_.nominees_id as nominees2_4_0_, nominee1_.id as id1_5_1_, nominee1_.employee_id as employee2_5_1_, nominee1_.nomination_id as nominati3_5_1_, employee2_.id as id1_0_2_, employee2_.email as email2_0_2_, employee2_.first_name as first_na3_0_2_, employee2_.last_name as last_nam4_0_2_, employee2_.manager_id as manager_6_0_2_, employee2_.nnumber as nnumber5_0_2_, employee3_.id as id1_0_3_, employee3_.email as email2_0_3_, employee3_.first_name as first_na3_0_3_, employee3_.last_name as last_nam4_0_3_, employee3_.manager_id as manager_6_0_3_, employee3_.nnumber as nnumber5_0_3_, nomination4_.id as id2_3_4_, nomination4_.user_id as user_id3_3_4_, nomination4_.nomination_type as nominati1_3_4_, user5_.id as id1_7_5_, user5_.employee_id as employee2_7_5_, employee6_.id as id1_0_6_, employee6_.email as email2_0_6_, employee6_.first_name as first_na3_0_6_, employee6_.last_name as last_nam4_0_6_, employee6_.manager_id as manager_6_0_6_, employee6_.nnumber as nnumber5_0_6_ from nomination_nominees nominees0_ inner join nominee nominee1_ on nominees0_.nominees_id=nominee1_.id left outer join employee employee2_ on nominee1_.employee_id=employee2_.id left outer join employee employee3_ on employee2_.manager_id=employee3_.id left outer join nomination nomination4_ on nominee1_.nomination_id=nomination4_.id left outer join user user5_ on nomination4_.user_id=user5_.id left outer join employee employee6_ on user5_.employee_id=employee6_.id where nominees0_.nomination_id=?