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如何在JPA中添加用户实体和好友关系实体的外键

2017-09-07 09:23:50  来源:网友分享

现有用户实体,相关代码如下:

@Entity@Table(name = "t_users")public class User extends Model {    @Id    @Column(name = "users_id")    @GeneratedValue(strategy = GenerationType.AUTO)    public Long userId;    ...}

以及好友关系实体,相关代码如下:

@Entity@Table(name = "t_relationship")public class Relationship extends Model {    @Id    @Column(name = "rs_id")    public Long relationshipId;    @Column(name = "rs_fromuser", nullable = false)    public User fromUser;    @Column(name = "rs_touser", nullable = false)    public User toUser;    ...}

从SQL书写来考虑,只需要在关系实体的两个Column上添加外键约束即可。但却不知道在JPA中应该怎么完成。

我个人数据库相关的知识比较欠缺,在这两个实体是不是多对多关系上存在疑问。
一个用户可以对应多个好友关系,一个好友关系对应固定的两个用户。应该说是n..2这种关系吗?

已经尝试使用@ManyToMany注解,但是我使用的Play 2框架和Ebean没有生成相对应的SQL语句。

@Entity@Table(name = "t_users")public class User extends Model {    @Id    @Column(name = "users_id")    @GeneratedValue(strategy = GenerationType.AUTO)    @ManyToMany(cascade = CascadeType.PERSIST, fetch = FetchType.LAZY)    @JoinTable(name = "users_relationship", joinColumns = {            @JoinColumn(name = ("user_id"), referencedColumnName = "userId"),            @JoinColumn(name = ("rs_id"), referencedColumnName = "relationshipId")    })    public Long userId;    ...}
@Entity@Table(name = "t_relationship")public class Relationship extends Model {    @Id    @Column(name = "rs_id")    public Long relationshipId;    @Column(name = "rs_fromuser", nullable = false)    @ManyToMany(targetEntity = User.class, mappedBy = "userId")    public User fromUser;    @Column(name = "rs_touser", nullable = false)    @ManyToMany(targetEntity = User.class, mappedBy = "userId")    public User toUser;    ...}

生成的SQL:

create table t_relationship (  rs_id                     bigint auto_increment not null,  #这里缺少了两个user id列  rs_makedate               datetime not null,  rs_accepted               tinyint(1) default 0 not null,  constraint pk_t_relationship primary key (rs_id));

补充:

后来注意到我需要映射的是User类而非User类的集合,在这里我大概是产生了一定的误解,需要搭配Play相关的内容来检查一下。

解决方案

2014.4.20再次更新

虽然上次的方法已经能够解决问题,但在开发过程中累积了一些经验后发现那样的解决方法算不上好。

下面更新下我现在的方案。

// User model类@Entity@Table(name = "t_users")public class User extends Model {    @Id    @Column(name = "u_id")    @GeneratedValue(strategy = GenerationType.AUTO)    public Long userId;    @OneToMany(fetch = FetchType.LAZY, mappedBy = "fromUser")    public List<Relationship> followUsers;    @OneToMany(fetch = FetchType.LAZY, mappedBy = "toUser")    public List<Relationship> followers;    ...}@Entity@Table(name = "t_relationship")public class Relationship extends Model {    @Id    @Column(name = "rs_id")    public Long relationshipId;    @JoinColumn(name = "rs_fromuser", nullable = false, updatable = false)    @ManyToOne(optional = false)    public User fromUser;    @JoinColumn(name = "rs_touser", nullable = false, updatable = false)    @ManyToOne(optional = false)    public User toUser;    ...}

以下为2014.3.18答案

问题经过探索后已经解决,但并没有完全理解其中的门道。关于JPA和ORM的基础知识不够扎实,希望能得到解惑。

就如前面所怀疑的,这并不是多对多关系,而是简单的一对一,不过是一个实体中有两个一对一关系罢了。

换用@OneToOne注解,并且了解了一下其他相关的内容,修改为下面的样子。

// User model类@Entity@Table(name = "t_users")public class User extends Model {    @Id    @Column(name = "users_id")    @GeneratedValue(strategy = GenerationType.AUTO)    public Long userId;    ...}// Relationship model类@Entity@Table(name = "t_relationship")public class Relationship extends Model {    @Id    @Column(name = "rs_id")    public Long relationshipId;    @JoinColumn(name = "rs_fromuser", referencedColumnName = "users_id", nullable = false, updatable=false)    @OneToOne(optional = false, targetEntity = User.class)    public User fromUserId;    @JoinColumn(name = "rs_touser", referencedColumnName = "users_id", nullable = false, updatable=false)    @OneToOne(optional = false, targetEntity = User.class)    public User toUserId;    ...}