Question

App 和 History 和OneToMany：

@Entity
@Table(name = "app")
@SuppressWarnings("serial")
public class App implements Serializable {
    @Id     @Column(name = "app_id")
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long appId;
    private String name;
    @OneToMany(fetch = FetchType.LAZY, mappedBy = "app", cascade = CascadeType.PERSIST)
    private Set<History> history = new HashSet<>();
    //get.set
}

@Entity
@Table(name = "app_history")
@SuppressWarnings("serial")
public class History implements Serializable {
    @Id
    @Column(name = "history_id")
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long historyId;
    @JoinColumn(name = "appId")
    @ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.REFRESH)
    private App app;
    //get.set
}

正在查看日志：

create table app (app_id int8 not null, name varchar(255), primary key (app_id))

create table app_history (history_id int8 not null, app_id int8, primary key (history_id))

alter table app_history add constraint FKeuk3km6r7oklj5xc1ecwvxmkm foreign key (app_id) 
references app

期望上面的行是

alter table app_history add constraint FKeuk3km6r7oklj5xc1ecwvxmkm foreign key (app_id) 
references app (app_id)

那么，当jpa尝试创建表时，为什么(app_id)丢失了？

注意：我在

mysql v8.0.12
mysql-connector-java v8.0.13
spring-boot v2.1.0.RELEASE
完整代码here：
日志为here

Answer 1

更改您的

@JoinColumn(name = "appId")

到

@JoinColumn(name = "app_id")

这应该有效。

Answer 2

这是问题所在：您使用的是MySQL数据库，但JPA使用的方言适用于PostgreSQL：

ConcurrentHashMap

我建议您在spring: main: banner-mode: "off" datasource: url: jdbc:mysql://localhost:3306/cifi3?useSSL=false&allowPublicKeyRetrieval=true username: tester password: tester jpa: database-platform: org.hibernate.dialect.PostgreSQLDialect show_sql: true hibernate: ddl-auto: create-drop format_sql: true data: rest: basePath: /repo中更改database-platform: org.hibernate.dialect.PostgreSQLDialect

jpa引用外部表而不是JoinColumn中的外部列

2 个答案: