java - Spring，在一个查询中向现有的多对多列表添加一列

Question

我有一个 Project 实体，它具有以下属性:

  @ManyToMany
  @JoinTable(
          name = "employee_projects",
          joinColumns = @JoinColumn(name = "project_id"),
          inverseJoinColumns = @JoinColumn(name = "employee_id")
  )
  private List<Employee> employees;

当我向 employees 添加另一个 employee 时，我正在执行以下服务方法。

  Employee addEmployeeToProject(long id, Employee employee) {
    Project project = findByProjectId(id);
    List<Employee> oldEmployees = project.getEmployees();
    oldEmployees.add(employee);
    project.setEmployees(oldEmployees);
    projectRepository.save(project);
    return employee;
  }

效果很好。但后来我实际上开始记录 SQL 请求，这就是实际发生的情况:

2019-12-17 18:45:52.644 DEBUG 14324 --- [nio-8080-exec-1] org.hibernate.SQL                        : delete from employee_projects where project_id=?
2019-12-17 18:45:52.646 DEBUG 14324 --- [nio-8080-exec-1] org.hibernate.SQL                        : insert into employee_projects (project_id, employee_id) values (?, ?)
2019-12-17 18:45:52.648 DEBUG 14324 --- [nio-8080-exec-1] org.hibernate.SQL                        : insert into employee_projects (project_id, employee_id) values (?, ?)
2019-12-17 18:45:52.648 DEBUG 14324 --- [nio-8080-exec-1] org.hibernate.SQL                        : insert into employee_projects (project_id, employee_id) values (?, ?)
2019-12-17 18:45:52.650 DEBUG 14324 --- [nio-8080-exec-1] org.hibernate.SQL                        : insert into employee_projects (project_id, employee_id) values (?, ?)
2019-12-17 18:45:52.650 DEBUG 14324 --- [nio-8080-exec-1] org.hibernate.SQL                        : insert into employee_projects (project_id, employee_id) values (?, ?)
2019-12-17 18:45:52.651 DEBUG 14324 --- [nio-8080-exec-1] org.hibernate.SQL                        : insert into employee_projects (project_id, employee_id) values (?, ?)
2019-12-17 18:45:52.651 DEBUG 14324 --- [nio-8080-exec-1] org.hibernate.SQL                        : insert into employee_projects (project_id, employee_id) values (?, ?)
2019-12-17 18:45:52.651 DEBUG 14324 --- [nio-8080-exec-1] org.hibernate.SQL                        : insert into employee_projects (project_id, employee_id) values (?, ?)
2019-12-17 18:45:52.652 DEBUG 14324 --- [nio-8080-exec-1] org.hibernate.SQL                        : insert into employee_projects (project_id, employee_id) values (?, ?)
2019-12-17 18:45:52.652 DEBUG 14324 --- [nio-8080-exec-1] org.hibernate.SQL                        : insert into employee_projects (project_id, employee_id) values (?, ?)
2019-12-17 18:45:52.652 DEBUG 14324 --- [nio-8080-exec-1] org.hibernate.SQL                        : insert into employee_projects (project_id, employee_id) values (?, ?)
2019-12-17 18:45:52.652 DEBUG 14324 --- [nio-8080-exec-1] org.hibernate.SQL                        : insert into employee_projects (project_id, employee_id) values (?, ?)
2019-12-17 18:45:52.653 DEBUG 14324 --- [nio-8080-exec-1] org.hibernate.SQL                        : insert into employee_projects (project_id, employee_id) values (?, ?)
2019-12-17 18:45:52.653 DEBUG 14324 --- [nio-8080-exec-1] org.hibernate.SQL                        : insert into employee_projects (project_id, employee_id) values (?, ?)
2019-12-17 18:45:52.653 DEBUG 14324 --- [nio-8080-exec-1] org.hibernate.SQL                        : insert into employee_projects (project_id, employee_id) values (?, ?)

所以他首先删除所有列(从给定项目中删除所有员工)。然后再次添加所有以前的+新的。我想如果我更换整个集合就会发生这种情况。但我只需要插入一个对象，这样为了节省时间，我可以编写自己的插入查询，这样它只插入我想要插入的一名新员工，而不是删除所有内容并再次添加所有内容。但是，除了编写自己的查询之外，还有更好的方法在一个请求/查询中完成此操作吗？

编辑:

如果我使用 native 查询，我可以只定义 INSERT 查询吗？ JPQL 中不可能吗？

Answer 1

当然。首先，首先在单个事务中完成所有这些操作，从而受益于对托管实体的修改自动持久化这一事实。无需调用 save()。

然后，了解 Java 传递对象的引用。如果您获得项目员工的集合，您不会获得该集合的副本。您会得到它的引用。因此，将员工添加到该集合中会有效地修改该项目的员工集合。无需在项目中再次设置集合。

最后，你所需要的就是

@Transactional
public void addEmployeeToProject(long id, Employee employee) {
    Project project = findByProjectId(id);
    project.getEmployees().add(employee);
}