【问题标题】:create entity from database wizard jpa 2.1 @ManyToMany, fix list not needed从数据库向导 jpa 2.1 @ManyToMany 创建实体,不需要修复列表
【发布时间】:2016-05-07 17:55:49
【问题描述】:

我有一些像 multimedia_feature 这样的连接表。

向导将在 Multimedia 类中创建一个 List 属性:

...
@JoinTable(name = "multimedia_feature", joinColumns = {
   @JoinColumn(name = "feature_oid", referencedColumnName = "oid")}, inverseJoinColumns = {
   @JoinColumn(name = "multimedia_oid", referencedColumnName = "oid")})
@ManyToMany
private List<Feature> featureList;
...

因为我只需要一个多对一关系(一个功能可以有多个多媒体文件),我将 media_oid 标记为唯一。 在此之后,向导创建其他 2 个表(我认为是多余的)

@Entity
@Table(name = "multimedia_feature")
@XmlRootElement
@NamedQueries({
    @NamedQuery(name = "MultimediaFeature.findAll", query = "SELECT m FROM MultimediaFeature m"),
    @NamedQuery(name = "MultimediaFeature.findByMultimediaOid", query = "SELECT m FROM MultimediaFeature m WHERE m.multimediaFeaturePK.multimediaOid = :multimediaOid"),
    @NamedQuery(name = "MultimediaFeature.findByFeatureOid", query = "SELECT m FROM MultimediaFeature m WHERE m.multimediaFeaturePK.featureOid = :featureOid")})
public class MultimediaFeature implements Serializable {

    private static final long serialVersionUID = 1L;
    @EmbeddedId
    protected MultimediaFeaturePK multimediaFeaturePK;
    @JoinColumn(name = "multimedia_oid", referencedColumnName = "oid", insertable = false, updatable = false)
    @OneToOne(optional = false)
    private Multimedia multimedia;
    @JoinColumn(name = "feature_oid", referencedColumnName = "oid", insertable = false, updatable = false)
    @ManyToOne(optional = false)
    private Feature feature;
...
...

@Embeddable
public class MultimediaFeaturePK implements Serializable {

    @Basic(optional = false)
    @NotNull
    @Column(name = "multimedia_oid")
    private int multimediaOid;
    @Basic(optional = false)
    @NotNull
    @Column(name = "feature_oid")
    private int featureOid;
    ...
    ...

最后它在多媒体类中添加了一个属性:

....
@OneToOne(cascade = CascadeType.ALL, mappedBy = "multimedia")
    private MultimediaFeature multimediaFeature;
....

因为我有很多加入类,我会避免创建所有这些类。 我可以手动创建属性吗,例如:

@JoinTable(name = "multimedia_feature",
           @JoinColumn(name"feature_oid", referencedColumnName = "oid")
)
    @OneToOne(optional = false)
    private Feature feature;

或者这会妨碍正确的持久性?

【问题讨论】:

    标签: java jpa-2.1


    【解决方案1】:

    看起来Multimedia类中的feature属性应该是@ManyToOne关系。 默认情况下,为多对多关系和单向一对多关系的映射创建连接表。 如果你想避免加入类,我认为你可以像这样使用@JoinTable 来映射要素类中的多媒体属性:

    @OneToMany
    @JoinTable(name = "multimedia_feature",
        joinColumns = @JoinColumn(name = "feature_oid"),
        inverseJoinColumns = @JoinColumn(name = "multimedia_oid") )
    private List<Multimedia> multimediaList;
    

    如果您确实需要与连接表的双向关系,映射将是这样的:

    public class Feature implements Serializable {
        @OneToMany(mappedBy="feature")
        private List<Multimedia> multimediaList;
        ...
    }
    
    public class Multimedia implements Serializable {
        @ManyToOne
        @JoinTable(name = "multimedia_feature",
                joinColumns = @JoinColumn(name = "multimedia_oid") ,
                inverseJoinColumns = @JoinColumn(name = "feature_oid"))
        private Feature feature;
        ...
    }
    

    或者您可以通过在多媒体表中引入一个连接列(如 feture_oid)来完全删除具有双向关联的连接表。以便您可以轻松映射多媒体类中的特征属性:

    @ManyToOne
    @JoinColumn(name = "feature_oid")
    private Feature feature;
    

    在 Feature 类中会是这样的:

    @OneToMany(mappedBy="feature")
    private List<Multimedia> multimediaList;
    

    【讨论】:

    • 我不能在多媒体中添加属性,因为这个类有其他类与其他类的连接表......如果我为每个类添加一个属性,我会有很多“空”值在分贝...我认为这不是一个好习惯,对吧?那么,第一种方式,多媒体课我要写什么?
    • @ManyToOne(由多媒体列表映射) ?
    • @Marco 如果您确实需要保留连接表和双向键,那么您可以使用 Multimedia 作为拥有方并在那里定义连接,并在另一端使用 mappedBy 属性,即特征。我调整了我的答案来演示。
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