
所以在我切换到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text / uri-List,但这不是我想要做的.
当我提交POST时,我收到204响应,但我只能在我的控制台上看到sql选择而不插入任何内容.
编辑:我使用Postman,但这是一个curl命令,它/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和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=?这是我的相关课程:
@Entitypublic class Developer { @ID @GeneratedValue @Column(name = "developerID") private Long ID; private String name; @OnetoMany(mappedBy = "developer",cascade = CascadeType.ALL) private List另外一个:
@Entitypublic 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如果我遗漏任何其他相关信息,请告诉我,我会提供.最佳答案如果你想保持双向,你似乎有两个选择:
>删除mappedBy =“developer”并让JPA使用连接表来管理一对多关系.
开发商:
@OnetoMany(cascade = CascadeType.ALL)@Jointable( name="DeveloperGame",joinColumns = @JoinColumn( name="dev_ID"),inverseJoinColumns = @JoinColumn( name="game_ID"))private List游戏:
@ManyToOne@Jointable( name="DeveloperGame",joinColumns = @JoinColumn( name="game_ID"),inverseJoinColumns = @JoinColumn( name="dev_ID"))private Developer developer;>如果您不想使用连接表,请删除mappedBy =“developer”并添加@JoinColumn(使关系两侧的连接列具有相同的列名称
开发商:
@OnetoMany(cascade = CascadeType.ALL)@JoinColumn(name = "devID")private List游戏:
@ManyToOne@JoinColumn(name = "devID")private Developer developer;我想知道这是否是设计或者这是Spring Data REST中的错误. 总结
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