jpa 复合主键并可与唯一键连接

Posted 2023-02-24

【中文标题】jpa 复合主键并可与唯一键连接

【英文标题】：jpa composite primary key and jointable with unique key

【发布时间】：2012-10-25 12:26:43

【问题描述】：

这里是 sql 表： 选项主键是名称+值

CREATE TABLE `option` (
  id int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(45) NOT NULL,
  `value` varchar(45) NOT NULL,
  PRIMARY KEY (`name`, `value`),
  UNIQUE KEY `id_UNIQUE` (`id`)
)

产品主键是增量id

CREATE TABLE `product` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(50) NOT NULL,
  PRIMARY KEY (`id`)
) 

一个产品有几个选项（由唯一键“id”引用）

CREATE TABLE product_option (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `id_product` int(11) NOT NULL,
  `id_option` int(11) NOT NULL,
  PRIMARY KEY (`id`),
  KEY `id_product` (`id_product`),
  KEY `id_option` (`id_option`),
  CONSTRAINT `FK_product_option` FOREIGN KEY (`id_product`) REFERENCES `product` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION,
  CONSTRAINT `FK_option_product` FOREIGN KEY (`id_option`) REFERENCES `option` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION
  )

在 java 方面，我已经以这种方式映射“选项”

@Embeddable
public class OptionId implements Serializable
    @Column(name="value")
    private String value;

    @Column(name="name")
    private String name;



@Entity @Table(name="option")
public class Option  

    @Column(name="id", unique=true)
    @GeneratedValue(strategy=GenerationType.IDENTITY)
    private Long id;

    @EmbeddedId
    private OptionId primaryKey;

    public OptionId getPrimaryKey() 
        return primaryKey;
    

    public void setPrimaryKey(OptionId primaryKey) 
        this.primaryKey = primaryKey;
    

以这种方式生产

@Entity
@Table(name="product")
public class Product 

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

    @Column(name="name")
    private String name

    @OneToMany(cascade=CascadeType.ALL, fetch=FetchType.LAZY)
    @JoinTable(
            name="product_option",
            joinColumns = @JoinColumn( name="id_product"),
            inverseJoinColumns = @JoinColumn( name="id_option", referencedColumnName="id")
    )
    private Set<Option> options;

然后在执行时我得到这个错误

Caused by: java.lang.ArrayIndexOutOfBoundsException: 1
    at org.hibernate.sql.SelectFragment.addColumns(SelectFragment.java:107)
    at org.hibernate.persister.collection.BasicCollectionPersister.manyToManySelectFragment(BasicCollectionPersister.java:308)
    at org.hibernate.persister.collection.BasicCollectionPersister.selectFragment(BasicCollectionPersister.java:294)
    at org.hibernate.loader.JoinWalker.selectString(JoinWalker.java:1070)
    at org.hibernate.loader.AbstractEntityJoinWalker.initStatementString(AbstractEntityJoinWalker.java:124)
    at org.hibernate.loader.AbstractEntityJoinWalker.initStatementString(AbstractEntityJoinWalker.java:109)
    at org.hibernate.loader.AbstractEntityJoinWalker.initAll(AbstractEntityJoinWalker.java:91)
    at org.hibernate.loader.AbstractEntityJoinWalker.initAll(AbstractEntityJoinWalker.java:78)
    at org.hibernate.loader.entity.CascadeEntityJoinWalker.<init>(CascadeEntityJoinWalker.java:52)
    at org.hibernate.loader.entity.CascadeEntityLoader.<init>(CascadeEntityLoader.java:47)
    at org.hibernate.persister.entity.AbstractEntityPersister.createLoaders(AbstractEntityPersister.java:3254)
    at org.hibernate.persister.entity.AbstractEntityPersister.postInstantiate(AbstractEntityPersister.java:3191)
    at org.hibernate.persister.entity.SingleTableEntityPersister.postInstantiate(SingleTableEntityPersister.java:728)
    at org.hibernate.impl.SessionFactoryImpl.<init>(SessionFactoryImpl.java:348)
    at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1872)
    at org.hibernate.ejb.Ejb3Configuration.buildEntityManagerFactory(Ejb3Configuration.java:906)
    at org.hibernate.ejb.HibernatePersistence.createContainerEntityManagerFactory(HibernatePersistence.java:74)
    at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.createNativeEntityManagerFactory(LocalContainerEntityManagerFactoryBean.java:225)
    at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.afterPropertiesSet(AbstractEntityManagerFactoryBean.java:308)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1477)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1417)
    ... 60 more

如果我删除“referencedColumnName”，我当然会得到错误

A Foreign key refering Option from Product has the wrong number of column. should be 2

但我不知道如何解决这个问题

【参考方案1】：

根据您的架构，您的实体注释略有偏差。在您指定“产品”和“选项”的地方，您应该使用“id_product”和“id_option”：

@OneToMany(cascade=CascadeType.ALL, fetch=FetchType.LAZY)
@JoinTable(
       name="product_option",
       joinColumns = @JoinColumn( name="id_product"),
       inverseJoinColumns = @JoinColumn( name="id_option", referencedColumnName="id")
)
private Set<Option> options;

顺便说一句，从长远来看，如果您在 option 表上切换主键和唯一键，您的生活将会轻松得多。

【讨论】：

我的错误。我已经纠正了错字。 （我仍然有问题）。出于商业目的，最好有一个有意义的主键，并且仅将技术 id 用于数据库目的

对，但在数据库级别，您应该处理“技术”ID。您的唯一密钥将成为您的“业务密钥”。今天晚些时候我会在工作中运行你的示例。看起来它应该按原样运行。