OneToMany 和 ManyToOne 映射 JPA / Hibernate

Question

我有两个具有外键关系的 table。我试过搜索如何做到这一点，它总是导致 OneToMany 和 ManyToOne 映射。我有这两个 tables.

user_role

用户

我正在尝试显示所有用户并通过将位置列的值获取到 user_role table.

来显示他们的位置

这些是我的类

User.java

@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;
    }

    .......
}

UserRole.java

@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;
    }

    .........

}

我一直收到这个错误。

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[=12=](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 所以我不是很熟悉。任何帮助将不胜感激。

Answer 1

您必须注释 Java 字段而不是 getter，例如：

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

和 User.class

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

另外，您在usertable中定义了role_id列吗？我在屏幕截图中看不到这个

Answer 2

你能把关系移动到像这样的相关对象吗

@ManyToOne()
@JoinColumn(name="role_id", referencedColumnName = "role_id", insertable = false, updatable = false)    
private UserRole userRole;

对用户角色也一样

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

更新

jar 文件似乎已损坏。尝试从 .m2\repository 和 mvn clean install

中删除内容

或

右键单击您的项目，select Maven，更新项目，勾选 Snapshots/Releases 的强制更新。

Answer 3

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。我在你上面描述的问题中看到你注释了字段 Id，然后注释了属性 userRole，如果我这样做，我会得到同样的错误。我可以通过仅注释属性或字段来修复错误。这两种模型都有效：

@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;
}
}

这也有效：

@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示例

@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;
}
}

测试代码（我添加这个是因为人们经常混淆必须设置双向关系才能让用户和 userRole 保存到 UserRole）。

    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());

Answer 4

提示，检查您的实体 classes 并确保您在 class 声明语句之前包含 @Entity 注释。

@Entity
public class MyEntityClass{
 //code
}

然后在@ManyToMany 或@OneToMany 关系下使用上述 MyEntityClass 的任何其他实体 class 添加以下内容。

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

这种方法对我有用，可能会有帮助。

Answer 5

你可以这样写一对多的关系：位置-外键

public class UserRole {
.....
@OneToMany(targetEntity = User.class,cascade =CascadeType.ALL)
@JoinColumn(name="position",referencedColumnName = "role_id")
private List<User> users; 
......  
}

Answer 6

创建和映射用户及其角色的另一种简单方法。定义两个实体 class 以及关系，它将创建第三个映射 table :

1.角色实体：

    @Entity
    @Table(name = "ROLE")
    @Data
    @AllArgsConstructor
    @NoArgsConstructor
    public class Role {
    
        @Id
        @GeneratedValue(strategy = GenerationType.AUTO)
        private long uuid;
        .
        .
        .
        
    }

2。用户实体：

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

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private long uuid;
    .
    .
    .
    @ManyToMany(cascade=CascadeType.ALL,fetch=FetchType.EAGER)
        @JoinTable(name="user_roles",
            joinColumns = {@JoinColumn(name="user_id", referencedColumnName="id")},
            inverseJoinColumns = {@JoinColumn(name="role_id", referencedColumnName="id")}
        )
    private List<Role> roles;
    
    
}

以上代码生成三个table :

角色 2.用户 3.USER_ROLE

OneToMany 和 ManyToOne 映射 JPA / Hibernate

OneToMany & ManyToOne mapping JPA / Hibernate

hibernate

jpa

one-to-many

many-to-one

spring-data-jpa