Question

我是JPA的新手，我在创建场景方面遇到了问题我有以下3个表。

"User" has UserId and other user Details. A Row in this is created when the User registers.  
"Event" has EventId and other Event Details. A Row is created when a new Event is created.
"Event_role_master" table has Role ID and other columns. Has Pre-loaded set 
of Roles.  
"Event-member" table has UserID,EventId,RoleId and TimeStamp.

我有一个场景，例如，当现有用户注册某个角色的可用事件时，应将UserId，事件ID，角色ID插入＆＃34; Event_Member表中。
我使用4th作为Mapping表在前3个表之间创建了多对多映射，但在大多数代码示例中，给出了当数据持久保存到User，Event，Role表时，它将在Event中创建一个条目 - 成员表本身但在我的情况下，我不需要插入前3个表，因为它已经包含数据。我只想从User，Event，Role表中获取并放入第4个表。我将使用JSON（REST服务）获取数据。有些人可以帮助或显示一些如何实现这种情况的代码示例。提前致谢。

Edit1：我们可以在JPA中使用继承策略吗。

    @Entity
    @Table(name="event_organiser_role")
    public class EventRole {

        @Id
        @GeneratedValue(strategy=GenerationType.AUTO)
        private Long id;
        private String role;
        private Timestamp created_at;
        private Timestamp updated_at;
        private String created_by;
        private String updated_by;
        private boolean is_deleted;

        @OneToOne(cascade=CascadeType.ALL)
        private PointMaster point;

        @OneToMany(mappedBy="role",cascade=CascadeType.ALL,orphanRemoval=false)
        private Set<EventMember> eventMember;


@Entity
@Table(name="user")
public class User {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;    
    private String name;    
    private String email;
    private String mobile;
    private int points;
    private Timestamp created_at;
    private Timestamp updated_at;
    private String created_by;
    private String updated_by;
    private boolean is_deleted;
    @Transient
    private Event event;
    @Transient
    private EventRole eventrole;

    @OneToMany(mappedBy = "user", cascade = CascadeType.ALL)
    private Set<UserBusiness> userBusiness;

    @OneToOne(mappedBy = "user", cascade = CascadeType.ALL)
    private UserAccount userAccount;

    @JsonIgnore
    @OneToMany(mappedBy="user",cascade=CascadeType.ALL,orphanRemoval=false)
    private Set<EventMember> eventMember;

@Entity
@Table(name="event_member")
public class EventMember implements Serializable{

    private static final long serialVersionUID = 3344138873871956378L;

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;
    private Timestamp created_at;
    private Timestamp updated_at;
    private String created_by;
    private String updated_by;
    private boolean is_deleted;

    @Id
    @ManyToOne
    @JoinColumn(name="event")
    private Event event;

    @Id
    @ManyToOne
    @JoinColumn(name="user")
    private User user;

    @Id
    @ManyToOne
    @JoinColumn(name="event_organiser_role")
    private EventRole role;

插入数据的方法（不工作）：

@Override
    public void registerEvent(User user){
        EventMember member=new EventMember();
        Event event=eventDAO.getEventByName(user.getEvent().getName());
        EventRole role=eventDAO.getRoleByName(user.getEventrole().getRole());
        user=(User) getUserbyValue(user.getEmail());
        member.setEvent(event);
        member.setRole(role);member.setUser(user);
        System.out.println(member);
        userDAO.addUser(user);
        eventDAO.createEvent(event);
        eventSer.createEventRole(role);
        //userDAO.registerEvent(member);
    }

在输入JSON中，只有用户电子邮件，Event_Name，Role-Name将在JSON中给出。所有这些都是这些表中的唯一键。

Answer 1

我认为以下是不正确的。

@Id
@ManyToOne
@JoinColumn(name="user")
private User user

1.对于所有加入的列，您不需要@ID 2.应更正为@JoinColumn（name =＆＃34; UserID）

@ManyToOne
@JoinColumn(name="UserID")
private User user

在数据库级别＆＃34; UserID＆＃34; ＆＃34;事件成员＆＃34; table应该是＆＃34; user＆＃34;的外键。表

相应地关注Event和EventRole

实现Spring数据JPA

1 个答案: