jpa复合主键和具有唯一键的jointable

时间:2012-11-05 16:14:28

标签: jpa composite-primary-key unique-key jointable

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`)
) 

产品有多个选项(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,我用这种方式映射“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;
}

然后在执行时我收到此错误

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”,我当然会收到错误

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

但我无法弄清楚如何解决这个问题

1 个答案:

答案 0 :(得分:0)

根据您的架构,您的实体注释略有偏差。如果您指定“产品”和“选项”,则应使用“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;

另外,如果您在选项表上切换主键和唯一键,从长远来看,您的生活会变得相当容易。