13

我有一个表，其中有两个外键指向两个不同的表，两个外键共享一列：

CREATE TABLE ZipAreas
(
  country_code CHAR(2) NOT NULL,
  zip_code VARCHAR(10) NOT NULL,
  state_code VARCHAR(5) NOT NULL,
  city_name VARCHAR(100) NOT NULL,
  PRIMARY KEY (country_code, zip_code, state_code, city_name),
  FOREIGN KEY (country_code, zip_code) REFERENCES Zips (country_code, code),
  FOREIGN KEY (country_code, state_code, city_name) REFERENCES Cities (country_code, state_code, name)
)

如您所见，有两个 FK 共享 country_code（巧合地在引用路径的末尾引用了同一列）。实体类看起来像（JPA 1.0 @IdClass）：

@Entity
@Table(name = "ZipAreas")
@IdClass(value = ZipAreaId.class)
public class ZipArea implements Serializable
{
    @Id
    @Column(name = "country_code", insertable = false, updatable = false)
    private String countryCode;

    @Id
    @Column(name = "zip_code", insertable = false, updatable = false)
    private String zipCode;

    @Id
    @Column(name = "state_code", insertable = false, updatable = false)
    private String stateCode;

    @Id
    @Column(name = "city_name", insertable = false, updatable = false)
    private String cityName;

    @ManyToOne
    @JoinColumns(value = {@JoinColumn(name = "country_code", referencedColumnName = "country_code"), @JoinColumn(name = "zip_code", referencedColumnName = "code")})
    private Zip zip = null;

    @ManyToOne
    @JoinColumns(value = {@JoinColumn(name = "country_code", referencedColumnName = "country_code", insertable = false, updatable = false), @JoinColumn(name = "state_code", referencedColumnName = "state_code"), @JoinColumn(name = "city_name", referencedColumnName = "name")})
    private City city = null;

    ...
}

如您所见，我将 countryCode 属性和城市的 country_code @JoinColumn 标记为只读（可插入 = 假，可更新 = 假）。Hibernate 因这句话而失败：

Exception in thread "main" javax.persistence.PersistenceException: [PersistenceUnit: geoinfo] Unable to configure EntityManagerFactory
    at org.hibernate.ejb.Ejb3Configuration.configure(Ejb3Configuration.java:374)
    at org.hibernate.ejb.HibernatePersistence.createEntityManagerFactory(HibernatePersistence.java:56)
    at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:48)
    at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:32)
    at tld.geoinfo.Main.main(Main.java:27)
Caused by: org.hibernate.AnnotationException: Mixing insertable and non insertable columns in a property is not allowed: tld.geoinfo.model.ZipAreacity
    at org.hibernate.cfg.Ejb3Column.checkPropertyConsistency(Ejb3Column.java:563)
    at org.hibernate.cfg.AnnotationBinder.bindManyToOne(AnnotationBinder.java:2703)
    at org.hibernate.cfg.AnnotationBinder.processElementAnnotations(AnnotationBinder.java:1600)
    at org.hibernate.cfg.AnnotationBinder.processIdPropertiesIfNotAlready(AnnotationBinder.java:796)
    at org.hibernate.cfg.AnnotationBinder.bindClass(AnnotationBinder.java:707)
    at org.hibernate.cfg.Configuration$MetadataSourceQueue.processAnnotatedClassesQueue(Configuration.java:3977)
    at org.hibernate.cfg.Configuration$MetadataSourceQueue.processMetadata(Configuration.java:3931)
    at org.hibernate.cfg.Configuration.secondPassCompile(Configuration.java:1368)
    at org.hibernate.cfg.Configuration.buildMappings(Configuration.java:1345)
    at org.hibernate.ejb.Ejb3Configuration.buildMappings(Ejb3Configuration.java:1477)
    at org.hibernate.ejb.EventListenerConfigurator.configure(EventListenerConfigurator.java:193)
    at org.hibernate.ejb.Ejb3Configuration.configure(Ejb3Configuration.java:1096)
    at org.hibernate.ejb.Ejb3Configuration.configure(Ejb3Configuration.java:278)
    at org.hibernate.ejb.Ejb3Configuration.configure(Ejb3Configuration.java:362)
    ... 4 more

老实说，这对我来说看起来很基本。“不允许在属性中混合可插入和不可插入的列”是一个如此弱的“借口”，不是吗？

Hibernate 是否应该能够处理这个问题，例如根据 JPA 规范？这是一个错误吗？

4

4 回答 4

31

有一种方法可以绕过验证并使其工作，从而表明该列是“@JoinColumnsOrFormulas”，然后提出解决方案：

错误：

@ManyToOne
@JoinColumns(value = {
    @JoinColumn(name = "country_code", referencedColumnName = "country_code"), 
    @JoinColumn(name = "zip_code", referencedColumnName = "code")})
private Zip zip = null;

@ManyToOne
@JoinColumns(value = {
    @JoinColumn(name = "country_code", referencedColumnName = "country_code", insertable = false, updatable = false),
    @JoinColumn(name = "state_code", referencedColumnName = "state_code"), 
    @JoinColumn(name = "city_name", referencedColumnName = "name")})
private City city = null;

好的：

@ManyToOne
@JoinColumns(value = {
    @JoinColumn(name = "country_code", referencedColumnName = "country_code"), 
    @JoinColumn(name = "zip_code", referencedColumnName = "code")})
private Zip zip = null;

@ManyToOne
@JoinColumnsOrFormulas(value = {
    @JoinColumnOrFormula(formula = @JoinFormula(value = "country_code", referencedColumnName = "country_code")),
    @JoinColumnOrFormula(column = @JoinColumn(name = "state_code", referencedColumnName = "state_code")),
    @JoinColumnOrFormula(column = @JoinColumn(name = "city_name", referencedColumnName = "name"))
})
private City city = null;

问候，

于 2012-10-30T20:48:55.733 回答
10

将支持 Hibernate 5，请参阅https://hibernate.atlassian.net/browse/HHH-6221

于 2012-03-05T14:01:19.133 回答
5

这在 Hibernate 5 中仍然没有解决。但是，如果我使用@JoinColumnsOrFormulas我会得到 ClassCastException。在所有连接列上附加insertable = false, updatable = false解决了我的问题：

例子：

@ManyToOne
@JoinColumns(value = {
    @JoinColumn(name = "country_code", referencedColumnName = "country_code", insertable = false, updatable = false),
    @JoinColumn(name = "state_code", referencedColumnName = "state_code", insertable = false, updatable = false), 
    @JoinColumn(name = "city_name", referencedColumnName = "name", insertable = false, updatable = false)})
private City city = null;
于 2017-11-15T18:32:34.277 回答
2

我正在使用休眠 5，但我仍然得到这个异常。如果您只将 insert="false", update="false" 添加到一个，您将收到一个异常，指出您混合了可插入和不可插入的列，并且这是不允许的。这是一个已经存在于跟踪器中的问题，但似乎没有得到解决。Hibernate 在多个重叠外键使用的列上抛出 AnnotationException

在我们的例子中，这意味着我们迁移到 EclipseLink，这实际上很容易，因为您主要需要替换 persistence.xml 并将 HSQL（Hibernate SQL）重写为 JPQL（JPA SQL）。此外，您可能需要替换自定义命名策略（Eclipse 将它们称为 SessionCustomizer）。当然，如果您使用休眠的特殊功能（例如休眠搜索等），可能会更难做到。但在我们的案例中，我们尝试修复重叠的外键数周，而最终迁移只需要数小时。

于 2016-07-07T14:27:16.680 回答