休眠的Java序列化automized數據庫

Question

假設我們有實體進入和EntryComment相關的多到一個如下圖所示

 @Entity 
     @Table(name = "entry") 
     public class Entry { 
      @Column(name = "author") 
      private String author; 
      @Column(name = "content") 
      private String body; 
      @Column(name = "comment_count") 
      private long commentCount; 
      @OneToMany(mappedBy = "entry") 
      private List<EntryComment> comments; 
     } 
    @Entity 
    @Table(name = "entry_comment") 
    public class EntryComment{ 
     @Column(name = "vote_count") 
     private long voteCount; 
     @ManyToOne 
     @JoinColumn(name = "id", referencedColumnName = "id", insertable = false, updatable = false) 
     private Entry entry; 
     @Column(name = "entry_id") 
     private long entryId; 
} 

兩個Entry和EntryComments有編號的。所以，有這樣的情況，其中entryId = 3213，它的評論ID是id = 3214和id = 42352。現在我收集EntryComments這樣：

Entry e = api.getEntryString(i); 
e = EntryService.save(e); 
for(int a = 0; a<e.getComments().size(); a++){ 
    EntryComment ec = EntryCommentService.save(e.getComments().get(a)); 
} 

這有可能在Hibernate中這樣評論序列直接從入口無「手動」遍歷集合？

Answer 1

爲了避免迭代EntryComments集合，我必須使用@Cascade註釋。據this應該

 @Entity 
     @Table(name = "entry") 
     public class Entry { 
      @Column(name = "author") 
      private String author; 
      @Column(name = "content") 
      private String body; 
      @Column(name = "comment_count") 
      private long commentCount; 
      @OneToMany(mappedBy = "entry") 
      @Cascade(CascadeType.SAVE_UPDATE) 
      private List<EntryComment> comments; 
     } 

，然後就

Entry e = api.getEntryString(i); 
e = EntryService.save(e); 

感謝，Riadh的暗示！