所以在切换到Boot之前这个工作正常。基本上我能够将text / uri-list发布到@OneToMany资源就好了。我将项目切换为使用Boot,并在此过程中停止工作。我可以在@ManyToOne结尾处输入一个text / uri-list,但这不是我想做的。
当我提交POST时,我得到204响应,但我只能在我的控制台上看到SQL选择而不是插入任何内容。 编辑:我使用的是Postman,但是这里有一个curl命令可以执行/返回相同的
curl -v -X POST -H "Content-Type: text/uri-list" -d "http://localhost:8080/games/2" http://localhost:8080/developers/1/gameList
IDEA上的记录器:
Hibernate: select developer0_.developer_id as develope1_1_0_, developer0_.name as name2_1_0_ from developer developer0_ where developer0_.developer_id=?
Hibernate: select game0_.game_id as game_id1_6_0_, game0_.developer_id as develope5_6_0_, game0_.esrb_rating as esrb_rat2_6_0_, game0_.name as name3_6_0_, game0_.release_date as release_4_6_0_, developer1_.developer_id as develope1_1_1_, developer1_.name as name2_1_1_ from game game0_ left outer join developer developer1_ on game0_.developer_id=developer1_.developer_id where game0_.game_id=?
以下是我的相关课程:
@Entity
public class Developer {
@Id
@GeneratedValue
@Column(name = "developerId")
private Long id;
private String name;
@OneToMany(mappedBy = "developer", cascade = CascadeType.ALL)
private List<Game> gameList;
其他人:
@Entity
public class Game {
@Id
@GeneratedValue
@Column(name = "gameId")
private Long id;
private String name;
private Date releaseDate;
private ESRBRating esrbRating;
@ManyToMany(mappedBy = "gameList", cascade = CascadeType.ALL)
private List<User> userList;
@ManyToOne
@JoinColumn(name = "developerId")
private Developer developer;
如果我遗漏了任何其他相关信息,请告诉我,我会提供。
答案 0 :(得分:2)
如果你想保持双向,你似乎有两个选择:
mappedBy = "developer"
并让JPA使用加入表来管理一对多关系。开发者:
@OneToMany(cascade = CascadeType.ALL)
@JoinTable(
name="DeveloperGame",
joinColumns = @JoinColumn( name="dev_id"),
inverseJoinColumns = @JoinColumn( name="game_id")
)
private List<Game> gameList;
游戏:
@ManyToOne
@JoinTable(
name="DeveloperGame",
joinColumns = @JoinColumn( name="game_id"),
inverseJoinColumns = @JoinColumn( name="dev_id")
)
private Developer developer;
mappedBy = "developer"
并添加@JoinColumn
(使您在关系的两边都有一个具有相同列名的连接列开发者:
@OneToMany(cascade = CascadeType.ALL)
@JoinColumn(name = "devId")
private List<Game> gameList;
游戏:
@ManyToOne
@JoinColumn(name = "devId")
private Developer developer;
我确实想知道这是设计还是Spring Data REST中的错误。
答案 1 :(得分:1)
Hibernate收集插入,更新和删除,直到刷新entitymanager。这通常在交易结束时完成。 所以可能是您的事务管理无法正常工作。
将org.springframework.transaction的日志记录设置为debug,而不是在打开和关闭事务时应该看到的。