Question

我有以下课程，我试图使用JPA获得onetomany关系。当xml消息发送到restcontroller时，消息存储在父表和子表中，但外键始终为null。如果有任何问题，请告诉我。

我用来保存数据的Rest服务，下面是我发布到这个rest方法的示例xml。

@RequestMapping(value="/team/", method = RequestMethod.POST , headers="Accept=application/xml")
@ResponseStatus(value = HttpStatus.OK)
public void createTeam(@RequestBody Team team) throws Exception {
    teamService.createTeam(team);
}

儿童班

   @XmlRootElement
    @Entity
    public class Player {

        @Id
        @GeneratedValue
        @Column(name = "player_id")
        private long player_id;

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

        @ManyToOne
        @JoinColumn(name = "team_id")
        private Team team;

and setter and getters....
}

和我的父类

   @XmlRootElement
    @Entity(name ="team")
    public class Team {

        @Id
        @GeneratedValue
        @Column(name = "team_id")
        private long team_id;

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

        @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "team")
        private List<Player> player;

and setter and getter
}

@Service
@Transactional
public class TeamServiceImpl implements TeamService{

protected EntityManager entityManager;

    public EntityManager getEntityManager() {
        return entityManager;
    }

    @PersistenceContext
    public void setEntityManager(EntityManager entityManager) {
        this.entityManager = entityManager;
    }

    @Override
    public List<Team> getTeam() {
         Query query = getEntityManager().createQuery("select a from team a");
          List<Team> resultList = query.getResultList();
          return resultList;
    }

    @Override
    public void createTeam(Team team) {
        getEntityManager().persist(team);
    }

}



@Service
@Transactional
public class PlayerServiceImpl implements PlayerService {

    protected EntityManager entityManager;

    public EntityManager getEntityManager() {
        return entityManager;
    }

    @PersistenceContext
    public void setEntityManager(EntityManager entityManager) {
        this.entityManager = entityManager;
    }


    @Override
    public List<Player> getPlayer() {
         Query query = getEntityManager().createQuery("select a from Player a");
          List<Player> resultList = query.getResultList();
          return resultList;
    }

    @Override
    public void createPlayer(Player player) {
        getEntityManager().persist(player);
    }

}

和我的xml

<team>
<name>CSK</name>
<Players>
<player>
    <name>kholi</name> 
</player>
<player>
    <name>king</name> 
</player>
<player>
    <name>raja</name> 
</player>
</Players>
</team>

以下是我在团队中创建条目后在表格中看到的结果。请注意播放器表中的team_id字段为空。这是我的问题。需要知道我如何修复它以使该字段获得team_id值。

mysql> select * from team;
+---------+-------+
| team_id | name  |
+---------+-------+
|       1 | Delhi |
+---------+-------+
1 row in set (0.00 sec)

mysql> select * from player;
+-----------+--------+---------+
| player_id | name   | team_id |
+-----------+--------+---------+
|         1 | kholi1 |    NULL |
|         2 | king   |    NULL |
|         3 | raja   |    NULL |
|         4 | kholi  |    NULL |
|         5 | a      |    NULL |
|         6 | b      |    NULL |
|         7 | c      |    NULL |
|         8 | d      |    NULL |
|         9 | e      |    NULL |
|        10 | f      |    NULL |
|        11 | g      |    NULL |
|        12 | h      |    NULL |
|        13 | i      |    NULL |
|        14 | j      |    NULL |
|        15 | k      |    NULL |
|        16 | l      |    NULL |
|        17 | m      |    NULL |
|        18 | n      |    NULL |
|        20 | sdsdsd |    NULL |
+-----------+--------+---------+
19 rows in set (0.00 sec)

Answer 1

看起来你已经保存了球员，但没有为他们设置球队。你应该这样做：

for (Player player: team.getPlayers())
    player.setTeam(team);

另一方面，只设置了一个玩家属性 - 它的名字。如果玩家在玩家收藏中而不是在团队中，则JPA无法自动检测到。

子表中的JPA外键为空

1 个答案: