JPA：堅持不插入到連接表

我正在使用JPA爲此應用程序和註釋映射實體。我有一個名爲UserStory的實體，另一個名爲Revision。有一個OneToMany供UserStory修改。

@Entity 
@Table(name = "user_story") 
@NamedNativeQueries({ 
    @NamedNativeQuery(name = "storyBacklog", query = "SELECT userstory.rank AS rank, userstory.description AS description, userstory.estimate AS estimate, userstory.name AS name, " 
     + "userstory.id AS id, userstory.status AS status FROM user_story userstory ORDER BY userstory.rank ASC", resultClass = UserStory.class), 
    @NamedNativeQuery(name = "getCos", query = "SELECT conditions.cos As cos FROM story_cos conditions WHERE conditions.story_id=?1", resultSetMapping = "cosMapping") }) 
@SqlResultSetMappings({ @SqlResultSetMapping(name = "cosMapping", columns = @ColumnResult(name = "cos")) }) 
public class UserStory implements Serializable { 

    private static final long serialVersionUID = 248298400283358441L; 

    @Id 
    @GeneratedValue(strategy = GenerationType.AUTO) 
    private Long id; 
    ... 
    @OneToMany(cascade = CascadeType.ALL) 
    @JoinTable(name = "story_revisions", joinColumns = @JoinColumn(name = "story_id"), inverseJoinColumns = @JoinColumn(name = "revision_id")) 
    private Set<Revision> revisions;

這裏的修訂實體：

@Entity 
@Table(name = "revision") 
public class Revision implements Serializable { 

    private static final long serialVersionUID = -1823230375873326645L; 

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

    @Column(nullable = false) 
    private String description; 

    @Column(name = "date_created", nullable = false) 
    @Temporal(TemporalType.TIMESTAMP) 
    private Date creationDate;

當我創建了一個用戶故事;我想補充一個版本上它，

，但是沒有填充連接表，除非我堅持的故事第一，再加入修訂和合並。

下面的代碼保存一個用戶故事：

public UserStory saveUserStory(UserStory userStory) { 
    Revision revision = new Revision(); 
    revision.setCreationDate(new Timestamp(System.currentTimeMillis())); 
    revision.setDescription("User story created"); 
    Set<Revision> revisions = new HashSet<Revision>(); 
    revisions.add(revision); 
    userStory.setRevisions(revisions); 
    return storyDao.create(userStory); 
    }

在StoryDao我所說的堅持方法：

@Transactional(readOnly = false) 
    public UserStory create(UserStory userStory) { 
    if (userStory.getRank() == null) { 
     Integer highestRank = 0; 
     highestRank = (Integer) entityManager.createNativeQuery("select max(rank) from user_story") 
      .getSingleResult(); 
     if (highestRank != null) 
     highestRank += 1; 
     else 
     highestRank = new Integer(1); 
     userStory.setRank(highestRank); 
    } 
    entityManager.persist(userStory); 
    LOGGER.debug("Added User Story with id " + userStory.getId()); 
    entityManager.detach(userStory); 
    return userStory; 
    }

這裏是從原木

Hibernate: 
    insert 
    into 
     user_story 
     (description, estimate, name, rank, status) 
    values 
     (?, ?, ?, ?, ?) 
Hibernate: 
    insert 
    into 
     revision 
     (date_created, description) 
    values 
     (?, ?) 
Hibernate: 
    select 
     revision0_.id as id5_0_, 
     revision0_.date_created as date2_5_0_, 
     revision0_.description as descript3_5_0_ 
    from 
     revision revision0_ 
    where 
     revision0_.id=? 
Hibernate: 
    select 
     userstory0_.id as id3_1_, 
     userstory0_.description as descript2_3_1_, 
     userstory0_.estimate as estimate3_1_, 
     userstory0_.name as name3_1_, 
     userstory0_.rank as rank3_1_, 
     userstory0_.status as status3_1_, 
     revisions1_.story_id as story1_3_3_, 
     revision2_.id as revision2_3_, 
     revision2_.id as id5_0_, 
     revision2_.date_created as date2_5_0_, 
     revision2_.description as descript3_5_0_ 
    from 
     user_story userstory0_ 
    left outer join 
     story_revisions revisions1_ 
      on userstory0_.id=revisions1_.story_id 
    left outer join 
     revision revision2_ 
      on revisions1_.revision_id=revision2_.id 
    where 
     userstory0_.id=?

我可以在SQL從這裏可以看到它保存了用戶故事和修訂，但是t在試圖插入連接表之前，母雞嘗試運行連接以查看關係是否存在。當然它不會找到，因爲我正在創建這個對象。

它在這種情況下如何獲得連接表？

來源

2011-10-19 Anadi Misra

現在工作。這裏的更新代碼

revisions.add(revision); 
userStory = storyDao.create(userStory); 
userStory.setRevisions(revisions); 
return storyDao.update(userStory);

我仍不確定爲什麼這是必需的;兩步法，我堅持一個對象然後更新它。

來源

2012-01-19 06:59:55

JPA：堅持不插入到連接表

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