播放2.3.4 Java - JPA oneToMany不保存外鍵。請檢查哪裏出錯

這裏是實體片段....... Ignore DaoObject ..它只是將自動生成的Id添加到它所屬的每個實體中。播放2.3.4 Java - JPA oneToMany不保存外鍵。請檢查哪裏出錯

我已經突出顯示了哪些內容需要保留。最終，它無法爲流表中的contentid列添加值。這雖然增加了一行內容和下面的代碼流表..

請檢查和幫助我解決問題

  Content newContent = TestHelper.contentFactory(null, "streamForCREATE_IT", xxxx, 100); 
      Stream stream = TestHelper.streamFactory(null, "name", "appname",); 
      **stream.setContent(newContent); 
      List<Stream> streams = new ArrayList<Stream>(); 
      streams.add(stream); 
      newContent.setStreams(streams);** 


@Entity 
public class Content extends DaoObject { 
    public Content() { 
    }; 

    private String name; 
    private ContentType type; 
    @OneToMany(fetch = FetchType.EAGER, mappedBy = "content", cascade = {CascadeType.PERSIST,CascadeType.MERGE}, orphanRemoval = true) 
    @JsonManagedReference 
    private List<Stream> streams = new ArrayList<Stream>(); 
    public String getName() { 
    return name; 
    } 
    public void setName(String name) { 
    this.name = name; 
    } 
    public List<Stream> getStreams() { 
    return streams; 
    } 
    public void setStreams(List<Stream> streams) { 
    this.streams = streams; 
    } 
    public void addStream(Stream newStream) { 
    if (streams == null) { 
     streams = new ArrayList<Stream>(); 
    } 
    newStream.setContent(this); 
    streams.add(newStream); 
    } 
} 

@Entity 
public class Stream extends DaoObject { 
    public Stream() { 
    } 
    @ManyToOne(fetch = FetchType.EAGER, cascade = CascadeType.ALL) 
    @JoinColumn(name = "contentid") 
    @JsonBackReference 
    private Content content;** 
    private String name; 
    private String appName; 
    public Content getContent() { 
    return content; 
    } 
    public void setContent(Content content) { 
    this.content = content; 
    } 
    public String getName() { 
    return name; 
    } 
    public void setName(String name) { 
    this.name = name; 
    } 
    public String getAppName() { 
    return appName; 
    } 
    public void setAppName(String appName) { 
    this.appName = appName; 
    } 
}

下面是MySQL工作臺的形象。

enter image description here