OneToMany＆amp; | 码农家园

OneToMany & ManyToOne mapping Spring Boot / Hibernate

我有两张有外交关系的桌子。我尝试过寻找如何做到这一点，它总是会导致一个Tomany&Manytoone映射。我有这两张桌子。

用户角色

enter image description here 。

用户

氧化镁

我试图通过将"职位"列的值获取到"用户角色"表来显示所有用户并显示他们的职位。

这些是我的课

用户.java

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24

@Entity
@Table(name="user")
public class User {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private long id;

..........

private UserRole userRole;

@ManyToOne()
@JoinColumn(name="role_id")
public UserRole getUserRole() {
return userRole;
}

public void setUserRole(UserRole userRole) {
this.userRole = userRole;
}

.......
}

用户角色.java

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28

@Entity
@Table(name="user_role")
public class UserRole {

.......

private List<User> user;

@OneToMany(targetEntity=User.class, mappedBy="userRole",cascade=CascadeType.ALL, fetch = FetchType.LAZY)
public List<User> getUser() {
return user;
}

public void setUser(List<User> user) {
this.user = user;
}

public long getRole_id() {
return role_id;
}

public void setRole_id(long role_id) {
this.role_id = role_id;
}

.........

}

号

我不断地得到这个错误。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38

org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'entityManagerFactory' defined in class path resource [org/springframework/boot/autoconfigure/orm/jpa/HibernateJpaConfiguration.class]: Invocation of init method failed; nested exception is javax.persistence.PersistenceException: [PersistenceUnit: default] Unable to build Hibernate SessionFactory; nested exception is org.hibernate.MappingException: Could not determine type for: java.util.List, at table: user_role, for columns: [org.hibernate.mapping.Column(user)]
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1699) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:573) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory.java:495) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
at org.springframework.beans.factory.support.AbstractBeanFactory.lambda$doGetBean$0(AbstractBeanFactory.java:317) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
at org.springframework.beans.factory.support.DefaultSingletonBeanRegistry.getSingleton(DefaultSingletonBeanRegistry.java:222) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
at org.springframework.beans.factory.support.AbstractBeanFactory.doGetBean(AbstractBeanFactory.java:315) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
at org.springframework.beans.factory.support.AbstractBeanFactory.getBean(AbstractBeanFactory.java:199) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
at org.springframework.context.support.AbstractApplicationContext.getBean(AbstractApplicationContext.java:1089) ~[spring-context-5.0.9.RELEASE.jar:5.0.9.RELEASE]
at org.springframework.context.support.AbstractApplicationContext.finishBeanFactoryInitialization(AbstractApplicationContext.java:859) ~[spring-context-5.0.9.RELEASE.jar:5.0.9.RELEASE]
at org.springframework.context.support.AbstractApplicationContext.refresh(AbstractApplicationContext.java:550) ~[spring-context-5.0.9.RELEASE.jar:5.0.9.RELEASE]
at org.springframework.boot.web.servlet.context.ServletWebServerApplicationContext.refresh(ServletWebServerApplicationContext.java:140) ~[spring-boot-2.0.5.RELEASE.jar:2.0.5.RELEASE]
at org.springframework.boot.SpringApplication.refresh(SpringApplication.java:780) [spring-boot-2.0.5.RELEASE.jar:2.0.5.RELEASE]
at org.springframework.boot.SpringApplication.refreshContext(SpringApplication.java:412) [spring-boot-2.0.5.RELEASE.jar:2.0.5.RELEASE]
at org.springframework.boot.SpringApplication.run(SpringApplication.java:333) [spring-boot-2.0.5.RELEASE.jar:2.0.5.RELEASE]
at org.springframework.boot.SpringApplication.run(SpringApplication.java:1277) [spring-boot-2.0.5.RELEASE.jar:2.0.5.RELEASE]
at org.springframework.boot.SpringApplication.run(SpringApplication.java:1265) [spring-boot-2.0.5.RELEASE.jar:2.0.5.RELEASE]
at com.rtc_insurance.AdminApplication.main(AdminApplication.java:10) [classes/:na]
Caused by: javax.persistence.PersistenceException: [PersistenceUnit: default] Unable to build Hibernate SessionFactory; nested exception is org.hibernate.MappingException: Could not determine type for: java.util.List, at table: user_role, for columns: [org.hibernate.mapping.Column(user)]
at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.buildNativeEntityManagerFactory(AbstractEntityManagerFactoryBean.java:402) ~[spring-orm-5.0.9.RELEASE.jar:5.0.9.RELEASE]
at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.afterPropertiesSet(AbstractEntityManagerFactoryBean.java:377) ~[spring-orm-5.0.9.RELEASE.jar:5.0.9.RELEASE]
at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.afterPropertiesSet(LocalContainerEntityManagerFactoryBean.java:341) ~[spring-orm-5.0.9.RELEASE.jar:5.0.9.RELEASE]
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1758) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1695) ~[spring-beans-5.0.9.RELEASE.jar:5.0.9.RELEASE]
... 16 common frames omitted
Caused by: org.hibernate.MappingException: Could not determine type for: java.util.List, at table: user_role, for columns: [org.hibernate.mapping.Column(user)]
at org.hibernate.mapping.SimpleValue.getType(SimpleValue.java:456) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
at org.hibernate.mapping.SimpleValue.isValid(SimpleValue.java:423) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
at org.hibernate.mapping.Property.isValid(Property.java:226) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
at org.hibernate.mapping.PersistentClass.validate(PersistentClass.java:597) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
at org.hibernate.mapping.RootClass.validate(RootClass.java:265) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
at org.hibernate.boot.internal.MetadataImpl.validate(MetadataImpl.java:329) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:461) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:892) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
at org.springframework.orm.jpa.vendor.SpringHibernateJpaPersistenceProvider.createContainerEntityManagerFactory(SpringHibernateJpaPersistenceProvider.java:57) ~[spring-orm-5.0.9.RELEASE.jar:5.0.9.RELEASE]
at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.createNativeEntityManagerFactory(LocalContainerEntityManagerFactoryBean.java:365) ~[spring-orm-5.0.9.RELEASE.jar:5.0.9.RELEASE]
at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.buildNativeEntityManagerFactory(AbstractEntityManagerFactoryBean.java:390) ~[spring-orm-5.0.9.RELEASE.jar:5.0.9.RELEASE]
... 20 common frames omitted

这是我第一次使用Spring，所以我不太熟悉。任何帮助都将不胜感激。

相关讨论

我试过了，但还是犯了同样的错误。我还将列名位置更改为role_id.：。(
可以在application.properties中设置spring.jpa.generate.ddl auto=create，让它为您创建表，然后让我知道是否仍然看到错误。
我使用了spring.jpa.hibernate.ddl auto=create。它仍然给我同样的错误。对不起，这是我第一次用弹簧。现在它给了我这个错误。
未能从[java. U.L.zip .zip文件中读取文件[ //Cu/Orac/Apg/Maven /Maven /Maven -/zWnJ；and Proj/2.2.1/Maven and zWnj.＆8203；Project 2.2.1.jar)：Java.UTIL.zIP.zIPExcExc:无效的LOC头(坏签名)。JA读取(原生方法)~[NA:1.80Y181] JAVA.UTI.zip .zIPFr.Access 1400美元(ZIPFr.java：60)~[Na:1.80Y181]
删除.m2
epository，然后运行mvn clean install
其他错误显示，所以我不得不安装几次。现在我的豆子出错了。
org.springframework.beans.factory.bean创建异常：创建类路径资源[org/springframework/boot/autoconfigure/orm/jpa/hibernatejpa&zwnj；&8203；configuration.class]&zwnj；&8203；中定义的名为"EntityManagerFactory"的bean时出错；嵌套异常为javax.persistence.persistence exception:[persistenceUnit:default]无法生成休眠会话工厂；嵌套的异常为org.hibernate.mappingeexception:repeated column in mapping for entity:com.rtc_Insurance.model.user column:role_id(应使用insert="false"update="fals"映射e")
无法生成Hibernate会话工厂；嵌套的异常为org.hibernate.mappingException:实体：com.rtc ou insurance.model.user列：role ou id映射中的重复列(应使用insert="false"update="false"进行映射)
是添加insert="false"update="false"它
嘿，非常感谢，错误已经消失了。但每次跑步都要丢掉桌子吗？以及如何在调用用户类时显示用户角色的标题。非常感谢。我已经试着修了3天了。
您可以先创建，然后再设置更新或不更新。不客气。如果我的回答对你有帮助，你能接受吗
您可以创建一个单独的问题，以了解如何从多到多获取特定数据，以便其他人也可以获得帮助，让我知道链接我将查看该链接，请解释您需要从表/实体中获得的确切问题。

我可以通过混合字段和属性之间的注释映射来重新创建您的异常。我在上面描述的问题中看到，您注释字段ID，然后注释属性userrole，如果我这样做，就会得到相同的错误。我可以通过只注释属性或字段来修复错误。这两种模式都可以工作：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39

@Entity
@Table(name ="USER")
public class User {

private Long id;
private UserRole userRole;
String userName;

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
public Long getId() {
return id;
}

public void setId(Long id) {
this.id = id;
}

@Column(name ="USER_NAME")
public String getUserName() {
return userName;
}

public void setUserName(String userName) {
this.userName = userName;
}

@ManyToOne()
@JoinColumn(name="role_id")
public UserRole getUserRole() {
return userRole;
}

public void setUserRole(UserRole userRole) {
this.userRole = userRole;
}
}

这也适用于：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41

@Entity
@Table(name ="USER")
public class User {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;

@ManyToOne()
@JoinColumn(name="role_id")
private UserRole userRole;

@Column(name ="USER_NAME")
String userName;

public Long getId() {
return id;
}

public void setId(Long id) {
this.id = id;
}

public String getUserName() {
return userName;
}

public void setUserName(String userName) {
this.userName = userName;
}

public UserRole getUserRole() {
return userRole;
}

public void setUserRole(UserRole userRole) {
this.userRole = userRole;
}
}

。

注释字段的其他模型对象的示例

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41

。

测试代码(我添加了这个代码，因为经常会混淆必须设置双向关系才能让用户和用户角色在保存到用户角色时保持不变)。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

List<User> users = new ArrayList<>();

User user1 = new User();
user1.setUserName("user1");
users.add(user1);

User user2 = new User();
user2.setUserName("user2");
users.add(user2);

UserRole userRole = new UserRole();
userRole.setRoleName("admin");
//Unidirectional relationship
user1.setUserRole(userRole);
user2.setUserRole(userRole);
//set Bidirectional relationship
userRole.setUsers(users);

userRole = userRoleRepository.save(userRole);

//Show that the two users and the UserRole persisted
UserRole result = userRoleRepository.findById(userRole.getId()).get();
assertEquals(2, result.getUsers().size());

只是一个提示，检查您的实体类，并确保在类声明语句之前包含@entity annotation。

1
2
3
4

@Entity
public class MyEntityClass{
//code
}

。

然后，在使用上述myEntityClass的任何其他实体类上，@manytomany或@onetomany关系下添加以下内容。

1
2
3

@OneToMany(mappedBy="mappingItem", cascade=CascadeType.ALL, orphanRemoval=true)
@JsonIgnore
private List<MyEntityClass> myEntityClassItemList;

这种方法对我有一定的帮助。

你必须注解Java字段，而不是GETTER。

1 2	@OneToMany(targetEntity=User.class, mappedBy="userRole",cascade=CascadeType.ALL, fetch = FetchType.LAZY) private List<User> user;

对于User.class。

1
2
3

@ManyToOne
@JoinColumn(name="role_id")
private UserRole userRole;

号

另外，您是否在user表中定义了role_id列？我在屏幕截图上看不到这个