Question

我正在尝试执行一些过滤，但是遇到了ManyToOne过滤的一些问题，并且每当我得到结果集时，我得到了所有Task对象，无论它们是由哪个项目链接。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
}

通过它的id与项目链接的任务类

@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 {

使用由org.hibernate.jpamodelgen.JPAMetaModelEntityProcessor生成的Metamodel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;
    }
}

服务电话：

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

不幸的是，每个查询都使用方法1和方法2，返回所有任务，无论项目ID是什么。我还启用了hibernate.show_sql，并在控制台上显示了以下内容

接近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 (? , ? , ? , ?))

接近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 (? , ? , ? , ?))

会非常满满的帮助。最诚挚的问候

Answer 1

您可能无法正确使用cb.and（）函数。此函数将返回一个新的组合谓词，而不是改变当前的谓词 所以而不是
cb.and(predicate, task.join(Task_.project).get(Project_.id).in(taskFilter.getProject().getId()));
你应该
predicate = cb.and(predicate, task.join(Task_.project).get(Project_.id).in(taskFilter.getProject().getId()));

适用于@ManyToOne的JPA规范CriteriaBuilder

1 个答案: