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Java 春天冬眠很多_Java_Spring_Hibernate_Jakarta Ee_Many To Many - Fatal编程技术网

Java 春天冬眠很多

Java 春天冬眠很多,java,spring,hibernate,jakarta-ee,many-to-many,Java,Spring,Hibernate,Jakarta Ee,Many To Many,我有两个表,视频,视频类别和加入表VIDEO2VIDEOCATEGORY。 我正在尝试进行多对多的注释。编译中没有错误或警告 数据库开头为空。我试图保持视频实体(与视频类别列表)…在这个时候都是正确的。新的视频被添加到数据库中。还添加了一些新类别。然后我尝试添加新的视频和新的独特的类别…正确的,以及。问题是当我尝试添加新视频时,DB中已经存在类别。在此尝试之前,我在VIDEO2VIDEOCATEGORY(“cat1”,3)和(“cat2”,3”)中有2条记录。尝试之后,应该有2条新记录-“cat

我有两个表,视频,视频类别和加入表VIDEO2VIDEOCATEGORY。 我正在尝试进行多对多的注释。编译中没有错误或警告

数据库开头为空。我试图保持视频实体(与视频类别列表)…在这个时候都是正确的。新的视频被添加到数据库中。还添加了一些新类别。然后我尝试添加新的视频和新的独特的类别…正确的,以及。问题是当我尝试添加新视频时,DB中已经存在类别。在此尝试之前,我在VIDEO2VIDEOCATEGORY(“cat1”,3)和(“cat2”,3”)中有2条记录。尝试之后,应该有2条新记录-“cat1”,4)和(“cat2”,4)以及2条现有记录(“cat1”,3)和(“cat2”,3),但旧记录不在DB中。他们被两张新唱片改写了

如何解决? 我的代码:



   import java.io.Serializable;
    import java.util.List;

    import javax.persistence.Column;
    import javax.persistence.Entity;
    import javax.persistence.FetchType;
    import javax.persistence.Id;
    import javax.persistence.JoinColumn;
    import javax.persistence.JoinTable;
    import javax.persistence.ManyToMany;

    import org.hibernate.annotations.Cascade;


    @Entity
    public class VideoCategory implements Serializable{

        private static final long serialVersionUID = -722840296120424003L;

        @Id 
        @Column(name = "id", unique = true, nullable = false)
        private String id;


        @ManyToMany(fetch=FetchType.EAGER)
        @Cascade(org.hibernate.annotations.CascadeType.SAVE_UPDATE)
        @JoinTable(name = "video2videocategory",
              joinColumns = {@JoinColumn(name = "cat_id")}, inverseJoinColumns =       @JoinColumn(name = "vid_id"))
          private List videos;


        public VideoCategory(){
        }

        public VideoCategory(String id) {
            super();
            this.id = id;
        }
        public String getId() {
            return id;
        }
        public void setId(String id){
            this.id = id;
        }
        public List getVideos() {
            return videos;
        }
        public void setVideos(List videos) {
            this.videos = videos;
        }

        @Override
        public String toString() {
            return "VideoCategory [id=" + id + ", videos=" + videos + "]";
        }
    }


    import java.io.Serializable;
    import java.util.Date;
    import java.util.List;

    import javax.persistence.Column;
    import javax.persistence.Entity;
    import javax.persistence.EnumType;
    import javax.persistence.Enumerated;
    import javax.persistence.GeneratedValue;
    import javax.persistence.GenerationType;
    import javax.persistence.Id;
    import javax.persistence.JoinTable;
    import javax.persistence.ManyToMany;
    import javax.persistence.SequenceGenerator;
    import javax.persistence.JoinColumn;

    import org.hibernate.annotations.Cascade;



    @Entity
    @SequenceGenerator(
        name="VID_SEQ_GEN",
        sequenceName="VIDSEQ",
        allocationSize=1
    )
    public class Video implements Serializable {

        private static final long serialVersionUID = 2284488937952510797L;

        @Id @GeneratedValue(strategy = GenerationType.SEQUENCE, generator="VID_SEQ_GEN")
        @Column(name = "id", unique = true, nullable = false)
        private Long id;

        @Column(name = "title", unique = false, nullable = false)
        private String title;

        @Column(name = "video_path", unique = false, nullable = false)
        private String videoPath;

        @Column(name = "video_type", unique = false, nullable = false)
        @Enumerated(value=EnumType.STRING) 
        private VideoType videoType;

        @Column(name = "creation_date", unique = false, nullable = false)
        private Date creationDate;

        @ManyToMany
        @Cascade(org.hibernate.annotations.CascadeType.SAVE_UPDATE)
         @JoinTable(name = "video2videocategory", 
                 joinColumns = {@JoinColumn(name = "vid_id")}, inverseJoinColumns = @JoinColumn(name = "cat_id"))
        private List categories;



        public Video(){
        }


        public Long getId() {
            return id;
        }
        public void setId(Long id) {
            this.id = id;
        }
        public String getTitle() {
            return title;
        }
        public void setTitle(String title) {
            this.title = title;
        }
        public String getVideoPath() {
            return videoPath;
        }
        public void setVideoPath(String videoPath) {
            this.videoPath = videoPath;
        }
        public VideoType getVideoType() {
            return videoType;
        }
        public void setVideoType(VideoType videoType) {
            this.videoType = videoType;
        }
        public Date getCreationDate() {
            return creationDate;
        }
        public void setCreationDate(Date creationDate) {
            this.creationDate = creationDate;
        }
        public List getCategories() {
            return categories;
        }

        public void setCategories(List categories) {
            this.categories = categories;
        }

        @Override
        public String toString() {
            return "Video [id=" + id + ", creationDate="
                    + creationDate + ", title=" + title
                    + ", videoPath=" + videoPath + ", videoType=" + videoType + "]";
        }
    }


我尝试了jpa和hibernate注释


请帮助。

错误可能是您将多对多关系定义为双方的父关系。 一边应该是孩子,像这样:

@ManyToMany(mappedBy = "categories") // in the class VideoCategory

显示您的
save()
方法似乎是在更新它,而不是创建新的。当您认为要添加
新视频时,请确保它实际上是视频的新实例,而不是现有实例。您的映射是正确的问题可能出现在
save()
中,或者更可能出现在您认为新的视频实例中,而实际情况并非如此。public void save(video video){getHibernateTemplate().save(video);}我正在创建一个新的视频实例,没有id,我尝试保存这个实例。