@ManyToOne的JPA规范标准生成器

Question

我正在尝试执行一些过滤，但ManyToOne的过滤有一些问题，每次我得到结果集时，我都会得到所有Task对象，而不管它们所链接的项目如何。我有以下实体：

@Entity
public class Project implements Serializable {

@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "project_seq_gen")
@SequenceGenerator(name = "project_seq_gen", sequenceName = "project_id_seq", allocationSize = 1)
private Long id;

@Column(unique = true)
private String projectId;

//more things
}

和任务类，它通过它的id与项目链接

@Entity
public class Task implements Serializable {

@Id
private String id;

@Column
private String name;

@ManyToOne(fetch = FetchType.EAGER)
@JoinColumn(name = "project_tasks")
private Project project;

@Column
private Enum<TaskType> type;

//more things....
}

在任务表中，它看起来像这样

 |id      |project_tasks | 
 |--------|--------------|
 | TASK1  |           1  |
 | TASK2  |           2  |
 |--------|--------------|

我有jparepository来搜索任务:

@Repository
public interface TaskRepository extends JpaRepository<Task, Integer> , JpaSpecificationExecutor {

以及使用由org . hibernate . jpamodelgen . jpametamodelentityprocessor生成的元模型的规范

 public class TaskSpecification implements Specification<Task> {
    private final TaskFilter taskFilter;

    public TaskSpecification(TaskFilter task) {
        this.taskFilter = task;
    }

    @Override
    public Predicate toPredicate(Root<Task> task, CriteriaQuery<?> cq, CriteriaBuilder cb) {
        Predicate predicate = cb.conjunction();
        //project
        //approach 1
        cb.and(predicate, task.join(Task_.project).get(Project_.id).in(taskFilter.getProject().getId()));  
        //approach 2
        cb.and(predicate, cb.equal(task.get(Task_.project).get(Project_.id), taskFilter.getProject().getId()));
        //some other filters to follow
        if (StringUtils.isNotBlank(taskFilter.getByState())) {
            if (TaskFilter.OPEN.equals(taskFilter.getByState())) {
                predicate = cb.and(predicate, task.get(Task_.state).in(TaskState.TO_DO, TaskState.ONGOING, TaskState.COMPLETE, TaskState.BLOCKED));
            } else {
                predicate = cb.and(predicate, task.get(Task_.state).in(TaskState.valueOf(taskFilter.getByState())));
            }
        }
        return predicate;
    }
}

服务呼叫：

public List<Task> findBySpecification(TaskFilter filter) {
    return taskRepo.findAll(new TaskSpecification(filter));
}

不幸的是，方法1和方法2的每个查询都返回所有任务，无论项目Id是什么。show_sql，控制台上显示了以下内容

方法 1

Hibernate: select task0_.id as id1_12_, task0_.active as active2_12_, task0_.task_assignee as task_as21_12_, task0_.create_date as create_d3_12_, task0_.description as descript4_12_, task0_.due_date as due_date5_12_, task0_.estimate as estimate6_12_, task0_.estimated as estimate7_12_, task0_.finishDate as finishDa8_12_, task0_.inSprint as inSprint9_12_, task0_.lastUpdate as lastUpd10_12_, task0_.loggedWork as loggedW11_12_, task0_.name as name12_12_, task0_.task_owner as task_ow22_12_, task0_.parent as parent13_12_, task0_.priority as priorit14_12_, task0_.project_tasks as project23_12_, task0_.release_id as release24_12_, task0_.remaining as remaini15_12_, task0_.state as state16_12_, task0_.story_points as story_p17_12_, task0_.subtasks as subtask18_12_, task0_.task_order as task_or19_12_, task0_.type as type20_12_ from Task task0_ inner join Project project1_ on task0_.project_tasks=project1_.id where 1=1 and (task0_.state in (? , ? , ? , ?))

方法 2

 Hibernate: select task0_.id as id1_12_, task0_.active as active2_12_, task0_.task_assignee as task_as21_12_, task0_.create_date as create_d3_12_, task0_.description as descript4_12_, task0_.due_date as due_date5_12_, task0_.estimate as estimate6_12_, task0_.estimated as estimate7_12_, task0_.finishDate as finishDa8_12_, task0_.inSprint as inSprint9_12_, task0_.lastUpdate as lastUpd10_12_, task0_.loggedWork as loggedW11_12_, task0_.name as name12_12_, task0_.task_owner as task_ow22_12_, task0_.parent as parent13_12_, task0_.priority as priorit14_12_, task0_.project_tasks as project23_12_, task0_.release_id as release24_12_, task0_.remaining as remaini15_12_, task0_.state as state16_12_, task0_.story_points as story_p17_12_, task0_.subtasks as subtask18_12_, task0_.task_order as task_or19_12_, task0_.type as type20_12_ from Task task0_ where 1=1 and (task0_.state in (? , ? , ? , ?))

我们会很乐意帮忙的问候

Answer 1

您可能无法正确使用cb.and()函数。这个函数将返回一个新的组合谓词，而不是改变当前的谓词。< br >所以不用< br> cb.and(谓词，task.join(Task_。项目)。get(Project_。id)。在(taskFilter.getProject()中。getId()))；
应该有< br> 谓词= cb.and(谓词，task.join(Task_。项目)。get(Project_。id)。在(taskFilter.getProject()中。getId()))；