java - 如何在 JPA 和 Hibernate 中保存列表的列表(List> 列表)

Question

我正在使用 Spring 服务器和 JPA 来保存变量。

我正在尝试保存:

@ElementCollection(fetch=FetchType.EAGER)
    List<List<String>> Listarespostas = new ArrayList<List<String>>();

但显然出现错误:

Exception in thread "main" org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'entityManagerFactory' defined in class palmaslab.mapas.config.Application: Invocation of init method failed; nested exception is javax.persistence.PersistenceException: [PersistenceUnit: default] Unable to build Hibernate SessionFactory
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1553)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:539)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory.java:475)
    at org.springframework.beans.factory.support.AbstractBeanFactory$1.getObject(AbstractBeanFactory.java:304)
    at org.springframework.beans.factory.support.DefaultSingletonBeanRegistry.getSingleton(DefaultSingletonBeanRegistry.java:228)
    at org.springframework.beans.factory.support.AbstractBeanFactory.doGetBean(AbstractBeanFactory.java:300)
    at org.springframework.beans.factory.support.AbstractBeanFactory.getBean(AbstractBeanFactory.java:195)
    at org.springframework.context.support.AbstractApplicationContext.getBean(AbstractApplicationContext.java:973)
    at org.springframework.context.support.AbstractApplicationContext.finishBeanFactoryInitialization(AbstractApplicationContext.java:750)
    at org.springframework.context.support.AbstractApplicationContext.refresh(AbstractApplicationContext.java:482)
    at org.springframework.boot.context.embedded.EmbeddedWebApplicationContext.refresh(EmbeddedWebApplicationContext.java:120)
    at org.springframework.boot.SpringApplication.refresh(SpringApplication.java:648)
    at org.springframework.boot.SpringApplication.run(SpringApplication.java:311)
    at org.springframework.boot.SpringApplication.run(SpringApplication.java:909)
    at org.springframework.boot.SpringApplication.run(SpringApplication.java:898)
    at palmaslab.mapas.config.Application.main(Application.java:54)
Caused by: javax.persistence.PersistenceException: [PersistenceUnit: default] Unable to build Hibernate SessionFactory
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.persistenceException(EntityManagerFactoryBuilderImpl.java:1225)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.access$600(EntityManagerFactoryBuilderImpl.java:119)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl$4.perform(EntityManagerFactoryBuilderImpl.java:853)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl$4.perform(EntityManagerFactoryBuilderImpl.java:843)
    at org.hibernate.boot.registry.classloading.internal.ClassLoaderServiceImpl.withTccl(ClassLoaderServiceImpl.java:399)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:842)
    at org.hibernate.jpa.HibernatePersistenceProvider.createContainerEntityManagerFactory(HibernatePersistenceProvider.java:150)
    at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.createNativeEntityManagerFactory(LocalContainerEntityManagerFactoryBean.java:336)
    at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.afterPropertiesSet(AbstractEntityManagerFactoryBean.java:318)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1612)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1549)
    ... 15 more
Caused by: org.hibernate.MappingException: Could not determine type for: java.util.List, at table: Project_Listarespostas, for columns: [org.hibernate.mapping.Column(Listarespostas)]
    at org.hibernate.mapping.SimpleValue.getType(SimpleValue.java:336)
    at org.hibernate.mapping.SimpleValue.isValid(SimpleValue.java:310)
    at org.hibernate.mapping.Collection.validate(Collection.java:315)
    at org.hibernate.cfg.Configuration.validate(Configuration.java:1362)
    at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1849)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl$4.perform(EntityManagerFactoryBuilderImpl.java:850)

有人知道替代方案吗？

Answer 1

开门见山，你根本做不到。

@ElementCollection 注释需要一个可嵌入类型的集合。这将被简单地转换为具有对父实体的 FK 引用的子表。

因此 Embeddable 映射子 SQL 表列。 List 不能是可嵌入的。

因为你不能嵌套可嵌入类型的集合:

An embeddable class (including an embeddable class within anotherembeddable class) that is contained within an element collection mustnot contain an element collection, nor may it contain a relationshipto an entity other than a many-to-one or one-to-one relationship. Theembeddable class must be on the owning side of such a relationship andthe relationship must be mapped by a foreign key mapping. (See Section2.9.)

您可以做的是创建一个新的实体类型 Child，它有一个 List 。

因此父子可嵌入列表变为:

@OneToMany
List<Child> children = new ArrayList<>();

Child 的样子:

@Entity
public class Child {

    @Id
    private Long id;

    @ElementCollection
    List<String> children = new ArrayList<>();
}